One. ff-& 




LLS« Ar^y , b\r CorpS> 



STRUCTURAL ANALYSIS 
AND DESIGN 

OF 

AIRPLANES 



Second Edition 



BY 

STRUCTURE AND AERODYNAMICS BRANCH 

AIRPLANE SECTION 

ENGINEERING DIVISION 



JA 



J 



■* 






LIBRARY OF CONGRESS 

RECEIVE" 
DOCUMENTS PWiSlON 




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Structural Analysis and Design 

OF 



Airplanes 



£/ 



ENGINEERING DIVISION 
McGOOK FIELD 
DAYTON, OHIO 



June, 1920 



Note 

The information contained herein is not to be 
republished, either as a whole or as extracts, with- 
out permission of the Chief of the Air Service. 



Published for the official information and guidance of the Air Service 
and those interested in aeronautics. 



By direction of the Chief of Air Service, 
Thurman H. Bane, Major, Air Service. 



PREFACE 
STRUCTURAL ANALYSIS AND DESIGN 

The purpose of this book is to give designers reasonable methods for 
the structural analysis and design of the component parts of the air- 
plane structure. The acceptance and use of a single system of analysis 
should tend towards uniformity in design, and ease in checking calcu- 
lations and in making comparisons between different airplanes. 

In the first chapter is given a brief development of the fundamental 
principles of graphical and analytical structural mechanics, and strength 
of materials, supplemented by detailed numerical examples. Under the 
head of "General Considerations," Chapter II deals with matters 
pertaining to an airplane cellule. In a brief discussion a few important 
features of structural design are considered, such as the location of in- 
terplane struts and spars, wing loading, aspect and gap-chord ratios, 
and stagger. Suggestions as to the choice of wing and strut sections 
are made. The work preliminary to the main analysis, that is, the dis- 
tribution of load between the wings, along the chord and on the wing 
tips, and the resolution of forces is explained. This is followed by con- 
cise statements of the general principles and methods of analysis, the 
detailed applications of which appear in the succeeding chapters. 

The general procedure followed in this book is to take as examples of 
the different structural units of an airplane, actual standard designs 
chosen to illustrate as many phases of the work as possible, and to fol- 
low through each step of the analysis and structural design. All the 
main equations and computations are given in full. These are supple- 
mented by explanations, drawings, and diagrams where necessary for 
clarifying the text. 

The appendix gives equations for the moments and deflections of 
continuous beams with various types of loading. For the more unusual 
cases the derivation of the equations is given. Data necessary for de- 
sign purposes, such as properties of strut sections, cable capacities, and 
ultimate strength of materials, are also placed in the appendix. 

To entirely separate stress analysis from design is impracticable, as 
the two are very closely correlated. For this reason a considerable 
amount of space has been devoted to a discussion of design in so far as 
it concerns structural problems. 

Three broad principles which are mentioned in the text deserve 
emphasis here. First, one of the two main purposes of any stress 
analysis is to enable a designer to so proportion structural members 
that there will be the correct "follow through," or that the strength of 
the members in one part of an airplane will bear the proper relation to 
the strength of the members in all its other elements. Second, wherever 
it is possible, and particularly in the larger types of airplanes, engineers 
should endeavor to design statically determinate structures even at the 
cost of a slight additional weight. There are always uncertainties in an 
indeterminate structure which occasionally cause serious failures. Third, 
every effort should be made to reduce secondary stresses. Sometimes 



these stresses become as large as the main stresses. Elimination of ec- 
centricities and, in the case of spars, avoidance of extremely long spans 
in which the deflections are great are two important means of minim- 
izing secondary stresses. 

Neither an economical design for a proposed airplane, nor a deter- 
mination of the strength of an existing structure closer than 10 to 20 
per cent can be made on the basis of approximate computations. Com- 
plete calculations for the structural members in an airplane require 
considerable time and care, yet it is being generally recognized that to 
restrict the structural engineering of airplanes is poor economy from the 
standpoint both of performance and safety. Furthermore, since a care- 
ful analysis can be depended upon to give an accurate prediction of 
the strength of a structure, the testing to destruction of full sized air- 
planes, especially the larger types, becomes less necessary. 

That the methods of analysis presented in this book are entirely sat- 
isfactory, or represent in any sense an ultimate development of the 
subject is not to be expected. They are, however, based in every case 
on sound engineering principles and practice, and may, therefore, be 
relied upon to give good results. The Engineering Division, basing its 
opinion on the stress analyses of the wing cellules of numerous airplanes 
which have been subjected to sand test, believes that the strength of 
a wing structure of conventional design and of a known quality of 
material can be predicted with an error of less than 5 per cent. The 
most difficult part of the work is not the calculation of the stresses so 
much as it is the determination of the ultimate allowable stresses in the 
material. Further experimental work will reduce the uncertainty in 
regard to some of these values. 



VI 



TABLE OF CONTENTS 



CHAPTER I 
Principles of Applied Mechanics and Strength of Materials 

Article Page 

1. Principles of Statics 1 

2. Law of Signs 1 

3. General Conceptions 1 

4. Characteristics of Reactions 2 

5. Determination of Reactions 3 

6. Method of Joints 4 

7. Character of Stress 5 

8. Method of Shear 5 

9. Method of Moments 5 

10. Methods of Moments and Joints 6 

11. Free Body Method 7 

12. Scope of Graphics 7 

13. Lettering of Truss 7 

14. Construction of Force Polygon 8 

15. Character of Stresses 9 

16. Development of Beam Theory 9 

17. Calculation of Moments and Shears 12 



Vll 



Contents 



Article Page 

18. Plotting of Moments and Shears 15 

19. Relations Between Moments and Shears 15 

20. Horizontal Shear 16 

21. General Equation 18 

22. Determination of Signs 18 

23. Combined Bending and Compression 22 

24. Column Formulas 23 

25. Wood and Steel in Combination 24 

26. Torsion, and Combined Torsion and Bending 26 

CHAPTER II 

General Considerations 

27. Visibility 28 

28. Choice of Wing Section 29 

29. Characteristics of Military Airplanes 41 

30. Structural Weight Analysis of Airplanes 41 

31. Load on Wings and Weight per Horsepower 41 

32. Aspect Ratio, and Shape of Wing Tips 44 

33. Location of Lift Wires in Inner Bay 45 

34. Location of Interplane Struts 47 

35. Location of Spars 50 

36. Torque of Engine 51 

37. Method of Least Work 52 

viii 



Contents 



Article Page 

38. Properties of Spar Sections 52 

39. Eccentricity of Fittings 53 

40. Maximum Load on Wings 54 

41. High and Low Incidence Conditions 55 

42. Static Test Conditions for the Wing Cellule 56 

43. Diving 58 

44. Distribution of Load Between Wings 60 

45. Distribution of Load Along Span and Chord 60 

46. Distribution of Pressure Along Wing Tip 61 

47. Distribution of Load Between Spars 61 

48. Resolution of Lift and Drag Forces 63 

49. Moments, Shears, and Reactions 63 

50. Truss Stresses 63 

51. Eccentric Strut Loads, and Wire Pulls 64 

52. Deflection Moments 64 

53. Stress Table 65 

54. Unit Fiber Stresses 65 



CHAPTER III 

Wing Stress Analysis 

55. Preliminary Data 70 

56. Location of Center of Pressure 70 

57. Distribution of Loads Between Spars 70 



IX 



Contents 



Article Page 

58. Distribution of Load Between Wings 70 

59. Load on Wings 75 

60. Distribution of Pressure on Wing Tips 75 

61. Load per Inch of Run 75 

62. Moments and Shears in Front Upper Spar 78 

63. Moments, Shears and Reactions 79 

64. Front Lift Truss 81 

65. Wing Spar Fittings 83 

66. Moments and Shears for Offset Strut Loads 83 

67. Moments and Shears for Eccentric Wire Pulls 85 

68. Summary of Moments, Shears and Reactions for Front 
Upper Spar 86 

69. Spar Properties 87 

70. Deflection of Front Lipper Spar 89 

71. Drag 89 

72. Drag Components of Struts and Wires 89 

73. Air Load Drag 95 

74. Loads on Drag Truss , . . . 95 

75. Struts , 95 

76. Stress Table 97 

77. Clark Truss 98 

78. S.P.A.D. Truss 100 

79. Thomas-Morse Truss 100 



X 



Contents 



Article Page 

80. S.V.A. Truss 100 

81. Monoplane Wings 100 

82. Multi-Spar Construction 103 

CHAPTER IV 

Miscellaneous Design 

83. Plywood Web Type 109 

84. Attachment of Rib to Spar 110 

85. Thick Sections Ill 

86. Split Capstrip 112 

87. Compression Ribs 112 

88. Truss Ribs 113 

89. Rib Loading 117 

90. Rib Spacing 118 

91. Streamline Section 119 

92. Hollow Circular Tubing 120 

93. Hollow Elliptical Tubing 122 

94. Design of Struts 122 

95. Combined Bending 125 

96. Taper of Struts 125 

97. Resistance of Struts 127 

98. Fineness Ratio 128 

99. Equivalent Weight 128 

xi 



Contents 



Article Page 

100. Center Section Struts 128 

101. Center Section Cross Wires and External Drift Wires 132 

102. Incidence Wires 133 

103. Built-up or Laminated Beams 134 

104- Splices in Spars and Longerons 137 

105. Bolts and Pins 138 

106. Lugs and Eye-bolts 139 

107. Horizontal vs. Vertical Bolts 140 

108. Pin Plates 141 

109. Wing Fitting 142 

CHAPTER V 

Airplane Chassis 

1 10. General Considerations 143 

111. Conditions of Loading 143 

112. Characteristics of the Chassis Structure 144 

113. Concerning the Chassis Used as an Example 145 

114. Kinds of Stresses to be Considered 147 

115. Calculation of Direct Stresses 147 

116. The Method of Least Work 152 

117. Distribution of Moments 153 

118. Resolution of Moments 158 

119. Actual Stresses and Moments in Members 159 

xii 



Contents 



Article Page 

120. Hinged Axle Chassis 165 

121. Problems in Chassis Design 167 

122. Design of the Axle 169 

123. Design of the Shock Absorber 169 

124. Factors of Safety 175 

125. Relation of Chassis to Fuselage 176 



CHAPTER VI 

Control Systems 

126. Introduction 177 

127. Calculation of Moments 177 

128. Investigation of Control Stick for Elevator Loads 180 

129. Investigation of Control Stick for Aileron Loads 183 

130. Investigation of Torque Tube to Which Control Stick is Se- 
cured 184 

131. Investigation of Aileron Lever 184 

132. Investigation of Bell Crank Lever 185 

133. Investigation of Teeth in Collar on Aileron Torque Tube . . 187 

134. Investigation of Aileron Torque Tube 188 

135. Investigation of Rudder Bar 188 

136. Investigation of Mast on Elevator Tubes 189 

137. Attachment of Mast to Tubes 190 

138. General Comments 193 

xiii 



Contents 



Article Page 

139. Investigation of Torque Tube Between Control Sticks 193 

140. Torque Tube Carrying Elevator Masts and Rear Stick 195 

141. Pursuit Airplane Control System 197 

142. Relation of Air Controls to Pilot 197 

143. Controls for Large Airplanes 197 

144. Ultimate Allowable Stresses 202 

CHAPTER VII 

Control Surfaces 

145. General 204 

146. Control Masts 204 

147. Bracing 205 

148. Loading 223 

149. Unbraced Surfaces 223 

150. Ribs 224 

151. Braced Surfaces 226 

CHAPTER VIII 

Fuselage 

152. Truss Type 228 

153. Monocoque Type 228 

154. Semi-Monocoque Type 229 

155. Airplane Considered for Analysis and Loading Conditions. . 229 

156. Computation of Loads for Flying Conditions 230 

157. Air Loads on Horizontal Tail Surfaces 234 

1 58. Traction and Torque 235 

xiv 



Contents 



Article Page 

159. Supporting Forces 236 

160. Stresses for Flying Conditions 237 

161. Flying Condition with Rudder Turned 237 

162. Landing with Tail Up 240 

163. Computation of Loads and Stresses for Three Point Land- 
ing Condition 243 

164. Summary of Stresses in Fuselage 245 

165. Computation of Factors of Safety 250 

166. Discussion of the Factors of Safety Computed 252 

167. Veneer Covered Truss Type Fuselage 253 

168. Semi-Monocoque Type 254 

169. Monocoque Type 255 

APPENDIX 

170. Prediction of High Speed at Ground 256 

171. Strength Factors for Wings, Fuselage and Tail Surfaces. . . . 256 

172. Equations for Continuous Beams 259 



173. Methods of Determining Deflection of Beams with Varying 
Load and Section 270 

174. Properties of Woods at 10 Per Cent Moisture 278 

175. Properties of Duralumin 289 

176. Miscellaneous Tables and Charts 292 

XV 



CHAPTER I 

PRINCIPLES OF APPLIED MECHANICS 

AND 

STRENGTH OF MATERIALS 

I. Applied Mechanics 

L Principles of Statics — The whole subject of statics, that part of 
mechanics dealing with bodies in equilibrium, is based on Newton's first 
law: "Every body continues in its state of rest or uniform motion in 
a straight line unless acted upon by an outside force." The corollary 
of this: "If any unbalanced force acts upon a body, the body is given an 
accelerated motion," states a condition when statical equilibrium no 
longer exists. Therefore, if a structure, or any part of a structure, is 
in equilibrium the algebraic sum of all the forces and of all the moments 
acting upon'it must equal zero. This may be briefly expressed by the 
following equations, 2 denoting "the sum of": %X — 0, ^Y = 0, 
2Z = 0, and %M = 0, in which X, Y and Z represent the components 
of the forces parallel to the X, Y and Z axes respectively, and M the 
moment of these forces about any point. As forces frequently lie in a 
single vertical plane they may be resolved into horizontal and vertical 
components. The fundamental equations of equilibrium then become: 
W — 0, %V = 0, and 2M = 0. 

2. Law of Signs — Care must be taken in problems to give each force 
or moment its proper sign. The following convention is in general use: 
forces acting to the right are positive, those to the left negative; forces 
acting upwards are positive, those acting downwards negative; clock- 
wise moments are positive, anti-clockwise moments negative. It should 
also be noted that a force has three characteristics: magnitude, direction, 
and point of application. 

The stresses in any statically determinate structure can be calcu- 
lated by various applications of the equations just given. Statically 
indeterminate structures are much more complicated. Two of the most 
useful methods for solving such cases will be discussed. For deter- 
minate structures two general methods of solution may be followed, 
either the analytical or the graphical. Frequently a combination of 
the two is most convenient. The analytical solution will be taken up 
first. 

THE SOLUTION OF A WARREN TRUSS 

3. General Conceptions — A truss is a structure, built up of members 
so arranged that they are subjected primarily to direct tension or com- 
pression, which transmits loads imposed upon it to the two or more 
points at which the truss is supported. If there are more than two 
supports the truss is termed continuous. This type will not be consid- 
ered here. Trusses are assumed to be in a single plane which is usually 
that of the forces acting upon the truss. In calculating the stresses in 

l 



Principles of Applied Mechanics 



Art. 4 



a truss it is assumed that the truss is "pin-jointed." By this is meant 
that at their points of intersection, the members are secured by a pin 
upon which they are free to turn. This condition seldom or never occurs 
in practice; the error that is made by the assumption is small. One 
important conception results from this assumption; namely, that any 
stress or load which goes into a member at a joint must act in the direc- 
tion of the axis of the member. No stress can be applied to a member 
except at a joint. The stresses acting at each end of a member must be 
equal, and may each be considered as a single force. Such a member 
is known as a "two-force member," as shown in Fig. 1. 



.^<£ 



Join? 



^* 



Fig. 1. Two Force Member 



4. Characteristics of Reactions — The first step in the solution of a 
truss is the determination of the known external loads upon the struc- 
ture. These should be shown upon a line sketch of the truss, together 
with their direction and point of application (see Fig. 2). When this 
is accomplished it is customary to solve for the supporting forces or 
reactions. The three equations of equilibrium, %V = 0, %H = 0, and 
2M = 0, which must be satisfied, enable us to learn three things about 
the reactions. As each of the two reactions has three characteristics; 
namely, magnitude, direction, and point of application, three facts, and 
only three, must be given if the structure is to be stable and statically 
determinate with respect to the external forces. Generally the points of 
application of both reactions, / and / 2 , and also the direction of one of 



p/n 




Fig. 2. Warren Truss with External Loads 

2 



Art. 5 Determination of Reactions 

them are known. Where rollers are used the reaction is always normal 
to the supporting surface. The conditions of equilibrium are sufficient 
to determine the magnitude of each reaction, and the direction of one. 
Should there be three reactions, or should another condition such as the 
direction of both reactions be unknown, the structure ceases to be 
statically determinate. If, on the other hand, only two unknown condi- 
tions regarding the reactions exist, the structure is unstable. It is im- 
portant at the outset to be able to tell whether the supporting forces 
can be calculated by statics. 

5. Determination of Reactions — The next step in computing the 
reactions is to take the moments of all the external forces acting upon 
the structure, including the reactions, about an axis passing through 
one of the reactions, usually the one whose direction is unknown. Since 
the algebraic sum of these moments equals zero, the reaction appearing 
in the equations may be found, as it is the only unknown. The hori- 
zontal and vertical components of the other reaction can now be ob- 
tained by resolving all the loads into their horizontal and vertical com- 
ponents, and by writing and solving the two equations, HH = and 
2F = 6>. A check on the calculation of the reactions is obtained by 
taking moments about the other reaction point. Incidentally, it is fre- 
quently much simpler to calculate the moments of the components of 
forces than of the forces themselves, especially if one of the components 
is eliminated by taking advantage of the fact that the moment of a force 
about an axis is the same for all points along the line of action of the 
force. An illustration of this is given in equation 4, in which the 
moments of the components h l and z\ of the left hand reaction r x are 
taken, rather than the moment of the reaction itself, which would be 
difficult to compute. In this manner the horizontal component, which 
passes through the axis of moments, is eliminated. In the same way it 
is much easier to obtain the moments of the two forces at u 2 than the 
moment of their resultant. 

CALCULATION OF REACTIONS FOR WARREN TRUSS 

(1) 5Mi = 600.5 + 1000-15— 400.10 — 20r r = 

r r — +700 lbs. 

(2) 2H = — 400 4-11! = 

h, — +400 lbs. 

(3) 2V = —600—1000 + 700 + v, = 

_ v, = +900 lbs. 
r, = V400 2 + 900 2 = +985 lbs. 
In Fig. 3 both r x and its components are shown, but, of course, the 
components take the place of the reaction itself. 
Check equation. 

(4) 2M l2 = 20.v 1 + 0.h 1 — 600.15 — 1000.5 — 400.10 = 

v, = +900 lbs. 
The solution for the stresses in the members may be either analytical 
or graphical. 

3 



Principles of Applied Mechanics 



Art. 6 



/000 




Fig. 3. Warren Truss and Reactions 



THE ANALYTICAL SOLUTION FOR THE TRUSS STRESSES 

6. Method of Joints — The three methods illustrated by this prob- 
lem, those of joints, of shear, and of movements, are but different appli- 
cations of the three equations of equilibrium. Consider first the joint l , 
Fig. 3, at which there are only two unknown stresses. So far as its 
equilibrium is concerned, the members l u x and l l ± may be cut, if they 
are replaced by the respective stresses acting in these members, as shown 
in Fig. 4. By applying ^V=0, in equation 5, the vertical component 
of the stress in l Q u x is obtained. The slope of this member gives the 
relation between v ± and h 1} and hence the value of h ± . Equation 7 gives 
directly the stress in l l x . This process is known as the method of 
joints. It is very simple and convenient in many cases, but for the more 
complicated joints it should be used in conjunction with the methods 
of shear and moments. No joint is capable of solution by this method 
at which there are more than two unknown stresses. The analytical 
method of joints is usually simplified by using the horizontal and vertical 
components of the known forces and unknown stresses. 

Joint 1 
Solution for stress in l Q u x and l 1 ± by the method of joints. 
(5) 2V = 900 + v lS =0 



-900 



(6) h x = vJ2 =—450 



The stress in 1 u ± ===== — V 4^0 2 _j_ 900 2 == — 1006 lbs. compression. 
(7) SH = 400 —450 + h 2 = 

50 lbs. tension. 



The stress in 1 l a 



h 2 = 



Art. 7 



Character of Stress 








<5W 








°'l 


3 


* "* 






1 








/4 

V 


400 A\\v 


m 


^ 




^ 


3 


JT<? 




9^ 









Fig. 4. Joint L 



Fig. 5. Truss to Left 
of Section B — B 



7. Character of Stress — In using the equations of equilibrium for 
calculating stresses the sign of the unknown stress or of its components 
should be written plus. Then the sign that is obtained upon solution 
of the equation determines the direction of the stress, whether toward 
or away from the joint, in accordance with the rules given in Art. 2. 
The nature of the stress in a member is thus automatically determined. 

8. Method of Shear — A process frequently used to obtain the stress 
in the web members of a truss where the chords are parallel is that of 
shear, illustrated by the solution for the stress in u ± l t . It consists sim- 
ply in taking a section through the truss which cuts the two parallel 
chords and the diagonal in question, and isolating part of the truss, as 
in Fig. 5. The condition for equilibrium of the vertical forces is now 
applied and, as v, is the only unknown vertical force, its value may be 
obtained. As before, h 3 is computed from v & . 

Section B — B 
Solution for stress in u^ l x bv the method of shear. 
(8) 2V==900 — 600 + v., = 

v.. =—300 

h 3 = v 3 /2 = +150 
Stress in u x \ 1 = +336 lbs. tension. 

9. Method of Moments — The condition for equilibrium of the hori- 
zontal forces ^H=0, Fig. 5, gives directly the stress in u x u 2 . This 
may also be calculated by the method of moments. In this method 
a section, B — B, is taken, and the part of the truss to the left of the 
section is isolated. All the members cut, except the one in which 
the stress is to be determined, must intersect, in this case at joint 
l ly Fig. 6. Equation 9 gives the moments of all the forces acting on the 

5 



Principles of Applied Mechanics 



Art. 9 




Fig. 6. Truss to Left of Section B— B 



part of the structure under consideration, including that of the stress 
in u x u 2 . This method is the most useful of the three explained, and 
should be thoroughly understood. 

Section B — B 
Solution for stress in u ± u 2 by the method of moments. 
(9) 5M 11 = 900.10— 600.5+10.h 4 = 
Stress in u x u 2 = h 4 = — 600 lbs. compression. 
. 10. Methods of Moments and Joints — As further examples of the 
method of joints and moments the solution for the stresses in l ± u 2 , u 2 / 2 , 
l x l 2 is given, the procedure being the same as before. 

Joint u 2 
Solution for stress in \ ± u 2 and u 2 1 2 by method of joints. 
(10 and 11) v 6 -=2h e ; v 5 = 2h 5 

(12) 2V = v 6 + v 5 — 1000 = 

(13) or 2h 6 + 2h 5 — 1000 = 

(14) 2H = h 6 + 600 — 400 — h 5 = 



Solving equations 13 and 14 simultaneously. 

h 6 = +150andh 5 = — 350 
v 6 = +300 and v 5 = +700 



Stress in \ ± u 2 = — Vl50 2 + 300 2 ^ — 336 lbs. compression 

— 789 lbs. compression 



u 2 l 2 



-V350 2 + 700 5 



Section D — D 
Solution for stress in l ± L by method of moments. 
(15) 2M U2 = — 700."5 + 10.h 7 = 
Stress in l x 1 2 = h 7 == +350 lbs. tension. 

6 



Art. 11 



Free Body Method 



/OOO 



600 \ c &a T 400 




600 w 


D 


/OOO 

' <?oo 


/ 


Bilk 


"l 


D 


700 



Fig. 7. Joint U 2 



Fig. 8. Truss to Right 
of Section D— D 



11. Free Body Method — It will be observed that in the solution just 
given the method employed in each case was the isolation of the struc- 
ture, in whole or in part, with all the external loads acting upon it, and 
the substitution for any members cut of the stresses in those members. 
This is a very general and important procedure applicable to nearly 
all problems in statics or kinetics, and is known as the "free body 
method." When a body is so isolated it is comparatively easy to tell 
whether it is statically determinate and whether it is in a condition of 
stable equilibrium. 

THE GRAPHICAL SOLUTION FOR THE TRUSS STRESSES 

12. Scope of Graphics — Any problem in statics that can be solved 
analytically is capable of a graphical solution. The graphical method 
is seldom used to obtain reactions. But often for stresses, particularly in 
a truss in which the web members have different slopes or the chords 
varying inclinations, it offers a ready solution to an otherwise tedious 
problem. This method is based on the proposition that, if the forces 
acting upon a structure are in equilibrium, they may be represented by 
the sides of a closed polygon. A force polygon is constructed by taking 
in order the forces acting upon a structure, and representing them, in 
both magnitude and direction, by the sides of the polygon. In Fig. 10, 
ABCDEFA forms the force polygon for the external forces applied to 
the truss. 

13. Lettering of Truss — The first step in the solution is to make an 
accurate line drawing (Fig. 9) of the structure, upon which all the ex- 
ternal forces are shown acting in the proper direction and at the correct 

7 



Principles of Applied Mechanics 



Art. 14 













/ooo 






,600 

1 X D 




£ 
4O0 






-600 






'/■ 


I 


V 1 


o \ 


400 




i-SO 




+JSO \ 


3 






Y " 







Fig. 9. Warren Truss for Graphical Solution 



Jca/e- / =S0 





F 

Fig 10. Stress Polygon 
for Warren Truss 



Fig. 11. Part of Stress Polygon 
Shown in Fig. 10 



location. Next, letter the truss in such a manner that on each side 
of every force and member there is one letter, and only one letter. 

14. Construction of Force. Polygon — The truss in Fig. 9 illustrates 
the correct method. Now start with one of the external forces, A — B 
and begin the construction of the force polygon for the external forces. 
In case all these forces are parallel, this polygon reduces to a straight 
line. Next, select a joint at which there are only two unknown members. 



Art. 15 Character of Stresses 

Take, for example, joint l , Fig. 2. The members C — / and / — A may 
be considered cut, and replaced by the stresses in them. The joint l 
must be in equilibrium under the action of the four forces, B — C, C — /, 
/ — A and A — B, which may therefore be represented by the sides of a 
polygon. This is constructed by drawing from point C, Fig. 11, a line 
parallel to the member C — / and indefinite in length, and from point A 
the line / — A, which is prolonged until it intersects line C — / at /. In 
the closed polygon thus formed C — / represents, in magnitude and direc- 
tion, the stress in member C — /, and / — A the stress in member / — A. 
In a similar manner, the stresses in the members D — H and H — / are 
found by considering all forces to the left of section X — Y. These too, 
being in equilibrium, may be represented by a polygon, which is 
A — B — C — / — H — D — C, of Figure 10. By the same method point 
G is located and the polygon completed. 

15. Character of Stresses — If each side of the polygon is scaled, the 
lengths of its sides give the stresses in all the members of the truss. To 
determine whether the stress in any member is tension or compression, 
read the letters which designate the member in a clockwise order with re- 
gard to the joint at either end of the member. Then, with the letters in 
the same order, read the stress in the member on the force polygon. If 
the stress, read in this manner, acts toward the joint, it is compression; 
if away from the joint, tension. For example: take members / — H 
and H — G, and use the joint l x at their intersection. The stress in 
/ — H is tension, because the line / — H in the polygon runs upward to 
the left away from the joint; that in // — G is compression, because the 
line H—G runs downward to the left toward the joint. 

II. Strength of Materials 

16. Development of Beam Theory — This subject is the division of 
mechanics which deals with the internal stresses in the members of a 
structure, and the manner in which the members resist the external 
forces acting upon them. The foundation for this entire section of me- 
chanics is the beam theory, which will be briefly developed. The pur- 
pose of the beam theory is to enable the stress at any point in any cross 
section of a rigid body subjected to bending to be calculated. The sim- 
plest case will be considered first. Let the beam in Fig. 12 be acted 
upon by a set of forces as shown. The forces and the beam have the 
following conditions imposed upon them: 

I. All the forces are parallel, and lie in a single, vertical plane 
passing through the axis of the beam and dividing each cross 
section symmetrically. 

II. The forces are perpendicular to the axis of the beam. 

III. The material of the beam is homogeneous. 

IV. The beam is of uniform cross-section throughout. 

Y. The axis of the beam is a straight line containing the center of 
gravity of every cross-section. (The term cross-section denotes 
a plane section perpendicular to the axis of the beam.) 

9 



Strength of Materials 



Art. 16 





^ a ? - 


G 




















a 3% 


f% 










as 


/9 


a 7 




F 












7 




■ 


i 


' 




t 


r 




X 








X 




* 














°'Ar 


o* 




1 i 
r 


F 








* \ ' 






\ 




'<?■ 


as 





Fig. 12. Beam with External Loads 







Section /?-/? 



Fig 13. Beam to Left of Section A — A 



In a beam conforming to these conditions, in shape, material, and 
loading, take a section A — A (Fig. 12), and isolate that portion of the 
beam to the left of the section (Fig. 13). Since the beam is in a condi- 
tion of equilibrium the internal forces acting upon the section must be 
such that the equations of equilibrium, %H = 0, %V = 0, and %M = 0, 
are satisfied. Since there are no external forces acting upon the beam 
in the direction of the Z — Z axis there can be no component of the in- 
ternal stress along this axis. That %V may equal zero a vertical com- 
ponent of the internal stress must exist so that f t — f 2 — / 3 — S = 
where S is the internal shear at the section. In the same manner the 
internal resisting moment of the components of the internal stress normal 
to the section must be such that f ± (x — a x ) — f 2 (x — a 2 ) — f 3 (x — a 3 ) + 
M = 0, where M is the internal resisting moment at the section. 
Furthermore, as none of the external forces have any horizontal com- 
ponent along the beam axis, the algebraic sum of these normal com- 
ponents of the internal stress must equal zero. They therefore form a 
couple. 

Consider any small element of area in the cross-section, AA, and let 
p T be the resultant internal stress acting upon it. p r may be resolved 
into two components, s, a shear stress, and p, a normal stress. 

10 



Art. 16 



Development of Beam Theory 



Then 3&A • s = S = 2,F, and %\A *p .y=M=%F (x—a), and 
%&A . p = 0,2, denoting "the sum of." 

There are three fundamental assumptions in the common theory of 
beams. 

I. A cross-section which is a plane before bending remains a plane 
after bending. 

II. "Hooke's Law" holds; that is, the stress is proportional to the 
strain throughout the beam. 

III. The ratio between the normal component of the stress at any 
point in a given cross-section to the strain in the direction 
parallel to the axis of the beam at that point is the same for 
both tension and compression, and has the same value that 
would be obtained from a bar of like material subjected to a 
uniform tension or compression. In other words the moduli 
of elasticity in tension and compression are equal. 

To make these assumptions perfectly clear the words stress and 
strain, as used in this connection, will be defined. Stress is the force of 
tension or compression acting upon a unit of area, and is expressed as 
pounds per square inch. Strain is the elongation or deformation per 
unit of length, or the total deformation divided by the length within 
which that deformation occurs; it is expressed in inches per inch. 
"Hooke's Law" merely states that until the unit stress reaches a certain 
value, known as the elastic limit of the material, the ratio between the 
stress and the strain is a constant quantity, called the modulus of elas- 
ticity, which is expressed in pounds per square inch. 

Consider now a portion of a bent beam (Fig. 14), and take two 
plane cross-sections A — B and C — D which before bending were a dis- 
tance L apart. If assumption / holds, some longitudinal layer mn 
will, after bending, have a length L 1 and the length of all other layers 




Fig. 14 Illustration of Beam Curvature 
11 



Strength of Materials Art. 17 

parallel to mn will be proportional to their distance from a fixed center. 
If the distance mn is taken short enough, the intersection of the neutral 
layer m with the plane of the forces can be considered as an arc of a 
circle of radius r. The length L x can be found from the proportion 

L x r + y 

— = . Let 8 equal the strain in the laver m x n x , and then 

L r 

L -\- h*L r 4- y y 

i> 1 =Z-[-8•^. From this = , or 8 = — . Hence, the 

L r r 

strains in the various layers of the beam are proportional to their dis- 
tances from the neutral layer. The intersection of the neutral layer 
with the plane of the forces is known as the neutral axis of the beam. 

It follows from assumptions II and III that the intensity of the 
normal component of the stress on any cross-section is zero at the neutral 
axis, and at any other point in the section is proportional to the distance 
of that point from the neutral axis. That is, the stress varies uniformly, 
and since the resultant is a couple, the neutral axis must pass through 
the center of gravity of the section. This variation in normal stress over 
the cross-section may be represented graphically, as in Fig. 19. 

If a equals the intensity of the normal stress at a unit's distance from 
the neutral axis, then the intensity of stress at a distance y equals 
p = ay. Substituting this value for p in the expression for M, 
5 A A *p*y, previously given, we have M = *% A A ^a^y 2 = a»^ A Ay 2 . 
Since I, the moment of inertia of the section, equals 25 A Ay 2 , M = al, 
=p»I/y. From this p = My/I, where p is the normal stress at any 
point at a distance 3; from the neutral axis, and M is the external bending 
moment at the section, p is called the "fiber stress." When y equals 
the distance from the neutral axis to the outside layer of the beam at 
the section of greatest external bending moment, p is a maximum, and is 
called the "maximum fibre stress." 

17. Calculation of Moments and Shears — The problem which fol- 
lows illustrates nearly all the points connected with the calculation of 
moments and shears, and the plotting of moment and shear curves. As 
the work is given in detail no further explanation is necessary. The 
usual graphical representation of bending moment and shear given in 
Fig. 18 will be found of help in determining the sign. 

The convention used in statics of calling clockwise moments plus 
and the reverse minus does not always hold. The standard, almost 
universally adopted, is that moments which produce compression in the 
top of a beam are positive and moments which produce tension in the 
top of a beam are negative. Shear is positive when the left half of the 
beam tends to slide up past the right; negative when the right tends to 
slide up past the left. These conventions of course apply not only to the 
total moment or shear at any point, but to the moment or shear caused 
by any individual load. 

The complete computations for the reactions, shears, and bending 
moments for a beam loaded as shown in Fig. 15 are as follows: 

12 



Art. 17 



Calculation of Moments and Shears 



CALCULATION OF REACTIONS 
125 2 X 1 



2M f 



5X70X9O—10OX125—50X80—90X55- 
200X15+^X100 = 



/OCr* ft? 


90* 

40' 


too* 


ic i 


D 


\ 


£ 


Uniform L'd =3* per /" 


yVf of beam - /*per '" 


1 ^5* Jf /OO" 


i 


A \kf 636* 




. ^* 


^77^ 



A 



Fig. 15. Beam with External Loads 




-4375 

Fig. 17. Moment Diagram 
13 



Strength of Materials 



Art. 17 



'G 



'enM/nn-^ 



r er>3/on- 



Posmi/e Momenf 

J-A7 




-m(\ 



Neyaf/ve Moment 



?<$ 



5/?n of Jheors 



Fig. 18. Graphical Representation of Sign for Moments and Shears 



£*. 




l/ar/at/'on of hor/z. 
5heor over 5ecf/on 

Fig. 19. Portion of Beam Showing Internal Bending and Shear Stresses 



ri = +638 lbs. = left reaction. 

2V= +638— 100— 50— 90— 200— 5X70— IX 125r r = 

r r = +277 lbs. = right reaction. 



Check- 



2M b = — 100X25+5X70X10+50X20+90X45+200X85 

+ 125X1X27.5— r r Xl00 = 0. 
r r = +277 lbs. = right reaction. 



CALCULATION OF SHEARS 

Shear at *— A = 
Shear at *+A = — 100 
Shear at — B = —100— (6X25) = —250 
Shear at +B = —250 + 638 = +388 

14 



Art. 18 Plotting of Moments and Shears 

Shear at — C = +388 — (20x6) = +268 

Shear at +C = +268 — 50 = +218 

Shear at — D = +218 — (6X25) = +68 

Shear at +D = +68 —90 = —22 

Shear at — E = —22 — (40X 1) = —62 

Shear at +E = —62 —200 = —262 

Shear at _F = — 262— (15xD =— 277 

Shear at +F = —277 + 277 = 

*+ designates the right side of section, — the left side. 

CALCULATION OF BENDING MOMENTS 

Moment at A — 

6X25 2 



Moment 


at 


B = 


100X25 


2 

6X45 2 

i 


= —4375 in. 


lbs. 










at 
at 


C = 

D = 


-100X45 - 
—200X40 - 


(638X20) = 

55 2 X1 

2 

f-4050 in. It 


= + 


?185 




lb- 


Moment 


2 
l- (277x55)- 


720 


in. 


lbs 


Moment 


at 


E = 


+277X15- 


15 2 X1 


)S. 









2 
Moment at F = 

18. Plotting of Moments and Shears — In constructing moment and 
shear diagrams it is convenient to remember that with concentrated 
loads only the shear is constant between loads and the moment varies 
as a straight line; with uniform distribution of loads the shear varies 
as a straight line and the moment as a parabola; with uniformly vary- 
ing distributed loads the shear curve is a parabola, while the moment 
curve is a cubic parabola. In all cases for downward loads the moment 
curve is convex upward for both a negative cantilever moment and the 
moment within the span. One very common mistake made both in 
plotting bending moments, and in using them in beam design, is the 
failure to convert moments expressed in foot-pounds into inch-pounds 
when the other units in the problem require this conversion. 

19. Relations Between Moments and Shears — Several extremely im- 
portant relations exist between the bending moment and shear in a 
beam. The bending moment at any point, or the ordinate to the moment 
diagram, is equal to the algebraic sum of the areas under the shear curve 
to that point multiplied by the scales to which the shear curve is con- 
structed. For example: referring to Figs. 15 and 17, the cantilever 

(.25 in.+.625 in.) 

moment at B = m b = X -833 in. X 30 X 400 = 

2 

— .364 sq. in. X 12,000 == — 4375 in. lbs.; or the moment at C = m c =- 

15 



Strength of Materials Art. 20 



X 12000 = +2185 in. lbs. As a corollary 



(.97 4- .67) 

-.364 H X -667 

2 

of this relation we have the fact that the maximum bending moment 
occurs at the point where the shear curve changes sign; that is, where 
the external shear is zero. (See Figs. 16 and 17.) Very often the only 
moment desired is the maximum one, and since the shear curve shows 
at once the one or two points where this moment can occur, the moment 
at these points can be computed without the necessity of drawing the 
moment curve. One more relation exists between the moment and shear 
that is much used. By it, it is possible to compute the moment at any 
section in a beam, knowing the moment at any other section, the shear 
at that section, and the loads between the two sections. The equation is: 
m x = m h + s h »x + ^F»a, %F»a being the moment about section X 
of all the loads between the sections. Whether the plus or minus sign 
is used with the second and third terms depends on the direction of the 
shear and the loads. If these are in such a direction as to cause a posi- 
tive bending moment, or one that would produce compression in the top 
of the beam, their sign is positive, otherwise it is negative. In this con- 
nection a positive shear can always be considered as an upward load, 
and a negative shear as a downward load. Example: 

6X45 2 

m d = — 4375 in. lbs. + 388 lbs. X 45 in. 50 lbs. X 25 in. = 

2 
+ 5720 in. lbs. (See Figs. 15-17.) 

20. Horizontal Shear — In the explanation of the theory of beam 
action the question of stresses due to shear was not touched upon. It was 
shown that the external shear at any section equals the algebraic sum 
of the forces on either side of the section. According to the last principle 
presented in the preceding paragraph there can be no increase in an 
external bending moment unless an external shear is present at the sec- 
tion over which this increase in moment occurs. Furthermore, in order 
that the internal resisting moment may increase, as the external moment 
becomes greater certain stresses are set up between the horizontal layers 
of a beam which are called "horizontal shear" stresses. Referring to 
Fig. 19, it is evident that if the bending moment at section B — B is 
larger than that at section A — A there will be a greater intensity of 
normal stress at all points on section B — B. If a portion of the beam 
above any horizontal plane, X — X, be isolated it will be seen that the 
condition of equilibrium for horizontal forces requires a horizontal shear, 
S, acting on the underside of the section equal to the difference between 
the normal forces on the ends. Letting b denote the breadth of the 
beam, and L, the distance between sections A — A and B — B , the in- 

S 

tensitv of horizontal shear on the X — X plane = s = . As shown in 

b.L, 
Fig. 19, the intensity of this shear stress has a parabolic variation from 

16 



Art. 20 



Horizontal Shear 



zero at the outside laver to a maximum at the neutral axis of the beam. 

S-0 
Its value is given by the formula, s = , in which S is the external 

b.I 
shear at the section, O the statical moment about the neutral axis of the 
portion of the beam above the horizontal plane (usually the one through 
the neutral axis) on which s is desired, b the breadth of the beam, and / 
the moment of inertia of the entire cross-section. 

Fig. 20 is a graphical representation of what takes place when in- 




Fig. 



Loo dec/ 

20. Illustration of Effect of Horizontal Shear 



stead of a single beam three separate planks are used. Because no 
resistance is offered to horizontal shear forces acting between the planks, 
they can not act as a unit, and the combination has but one-third the 
strength of a single beam of the same total depth and width. 

Beams must always be investigated for their strength in horizontal 
shear, for failure will occur just as readily from this cause as from excess 
of normal stress, if the beams are improperly designed. The example 
which follows makes the application of the shear formula clear. 

CALCULATION OF HORIZONTAL SHEAR STRESSES 
IN A BEAM 

Assume a beam of the section shown in Fig. 21. 

2X (3.6) 3 (2.0— .6) X (2.4) a 

Moment of inertia = I = = 6.17 

12 12 

Maximum statical moment = Q = 2X-6Xl-5 + .6XL2X-6 = 2.24 
External shear on the section = 1000 pounds. 
Maximum intensity of horizontal shear = s. 
1000X2.24 

— = 605 lbs. per sq. in. 



.6X6.17 



r 



Strength of Materials 



Art. 21 




Fig. 21. Section of a Beam 

In addition to horizontal shear stresses, stresses due to vertical shear 
are present. These stresses act in planes at right angles to the horizontal 
shear stresses, and at any point are equal to the latter, in accordance 
with the principle that shear stresses on any two planes at right angles 
to each other are equal in intensity. As a rule vertical shear stresses 
are not a limiting factor in design work. 



THE THEOREM OF THREE MOMENTS 

21. General Equatio?i — The derivation of the three moment equation 
from the theory of the elastic curve will not be taken up here. In the 
form in which it is used for uniformly distributed downward loads on 
a beam, with its supports either all on the same level or lying in the 
same straight line, the equation is as follows 



m^i 



2m 2 (l t + 1 2 ) 



m 



i.- + 



W-J 



w. 



+ 



4 4 

22. Determination of Signs — The determination of the sign of the 
shears in the solution of the three moment equation is apt to cause more 
confusion than any other part of the problem. A definite rule can be 
given, however, that will always hold whatever the type of loading. If 
the portion of the beam between a reaction point and a section at which 
the shear is to be determined tends to slide up by the part of the beam 
on the other side of the section, the shear is positive. If the tendency to 
slide is the reverse, the shear is negative. Fig. 18 is a graphical repre- 
sentation of this law. When first calculating the shears from the bend- 
ing moments at the reaction points by the equation, m 2 =m 1 +J - ; 1 »X+ 
%F *a, the sign of the shear term should always be written plus. If the 

18 



Art. 22 Determination of Signs 

solution of the equation gives the shear a positive sign, the shear is 
positive. But if the solution results in a negative sign, the shear is neg- 
ative, and must, therefore, be used with a minus sign in any further 
computations. It will be observed that a positive shear always tends to 
produce a positive moment, or one that causes compression in the top 
of a beam, while a negative shear tends to produce a negative moment. 
Attention should again be called to the fact that downward loads are 
negative, upward loads positive. In the right member of the general 
equation given above, both terms have a positive sign, but in the com- 
putations which follow, these terms become negative because w 13 and 
w 2 are negative. 

The solution for a typical case (See Fig. 22) is given below: 

MOMENTS 

10 X 25 2 10 X 20 2 
(1) m 1 = = — 3125 in. lbs. and m 4 = = 



2 2 

—2000 in. lbs. 

10 

(2) —3125 X 100 + 2m.,(100+50)+m..XSO = (100 :i +50 3 ) 

4 

(3) m 2 X300 + m.,x50 = — 2,500,000 

10 

(4) m.,X50 + 2m, (50+110) -2,000X110= (50 :; +110 3 ) 

4 

(5) m 2 X50 + m,X320 = —3,480,000 

Solve equations (3) and (5) simultaneously — 
m 2 = — 6,695 in. lbs; m 3 = — 9,835 in. lbs. 

SHEARS AND REACTIONS 

(6) SV = — 25X10 + s_ 1 = 0; s^ — +250 lbs. 

100 2 X10 

(7) —6695 =—3,125 + 100Xs +1 



—6695+3125 100 2 XlO 

(8) s+1 = ; = +464.3 lbs. 

100 2X100 

(9) r l=s . 1 +s +1 = +714.3 lbs. 

(10) SV = — 10X 100+464.3 +s_ 2 = 0;s_ 2 = +535.7 lbs. 

50 2 X10 

(11) —983 5 = — 6695+50s +2 



19 



Strength of Materials 



Art. 22 






<u 




PQ 




X 




3 




C 




3 




C 












C 


in 


O 


C 


{ ) 


O 






o 


73 






£ 
ed 


-o 


s~ 


e 

03 


03 


cn 


Q 


-o 








o 


03 


J 


_c 


-fl 


cn 


.ti 


T3 


< 


C 




w 






£ 


c 


cd 


a» 


cu 
PQ 


£ 
o 


3 


s 


O 


"oc 


3 
C 


C 


•rt 


■a 


G 


c 





a> 


u 


PQ 




^ 




<u 


a 


£ 


a 


o 


D J 


- — 


— ■ 


r\i 


ro 


(>i 


<\J 



20 



Art. 22 Determination of Signs 

(—9835+6695) 50 2 XlO 

(12) s +2 = 1 =+187.2 lbs. 

50 2X50 

(13) r 2 = s. 2 +s +3 = +732.9 lbs. 

(14) 2V = — 50X10+ 187.2 + s. 3 =0; s_ 3 = +312.8 lbs. 

110 2 X10 

(15) —2000 = — 9835 + 110 s,. 

2 

(_2000+9835) 110 2 X10 

(16) s +8 = 1 =+621.2 lbs. 

110 2XH0 

(17) r 3 = s. 3 +s +3 = +934.0 lbs. 

(18) SV = —110X10 + 621.2 + s_ 4 = 0; s. 4 = +478.8 lbs. 

(19) s +4 = +20X10= +200 lbs. 

(20) r 4 = s. 4 + s +4 = +678.8 lbs. 

CHECK 

(21) Total load= 10(25 + 100+50+110+20) = ri +r.,+r 3 +r 4 = 

3050 lbs. 

POINTS OF ZERO SHEAR 

a. a 464.3 

Span 1-2 X = — = = 46.4 in. from r, 

w 10 

187.2 

Span 2-3 X = = 18.7 in. from r., 

10 

621.2 

Span 3-4 X =■- =62.1 in. from r.. 

10 

CENTER MOMENTS 

46.4 2 X10 

(22) m,., = —3125 h ^6.4 X 464.3 = +7660 in. lbs. 



(23) m 2 _3 = — 6695 h 187.2x18.7 = —4940 in. lbs. 



2 

18.7 2 X10 


2 
62.1 2 X10 



(24) m 3 . 4 = — 9835 h 621.2x62.1 = +9460 in. lbs. 



The methods of solution used above are but applications of the 

21 



Strength of Materials 



Art. 23 



principles previously discussed, especially the relationship between the 
moments at two sections of a beam, the shear at one of the sections, and 
the loads between the sections. In determining the maximum moments 
in a span, the point of zero shear at which, as has been explained, this 
moment occurs, was first calculated. As in the present example, there 
may be two or more unknown moments over the supports, making it 
necessary to write the three moment formula as many times as there are 
unknown moments, and to solve the resulting equations simultaneously. 
It is first applied to the first and second interior spans at the left end of 
the beam, then to the second and third, and so on until the last interior 
span at the right end of the beam is reached. In the general equation 
m] is always the moment at the left end of the left of the two spans 
being considered, m 2 that between the spans, and m z that at the right 
end of the right span. 

For purposes of plotting, the law of signs given above must be slightly 
modified. If that portion of the beam to the left of a section !ends to 
slide up by the portion to the right of the section, the shear is positive. 
When the tendency to slide is the reverse, the shear is negative. This 
rule is illustrated by the part of Fig. 18 to the right of the reaction point. 

23. Combined Bending and Compression — When the forces acting 
on a beam do not conform in direction and point of application to the 
conditions stated at the beginning of the discussions of the beam theory, 
the formula for simple beam action, f=M •y/I, must be supplemented. 
In Fig. 24 none of the loads are both perpendicular to the axis of the 




Fig. 24 



beam and in a vertical plane passing through the neutral ^xis. The first 
step is to resolve all forces which are neither parallel nor perpendicular 
to this axis into two^ or, if necessary, three components which are either 
parallel or perpendicular to it. Those components or forces which are 
at right angles to the neutral axis of the beam produce bending about 
that axis of the cross-section to which they are perpendicular; those com- 
ponents or forces parallel to the neutral axis cause axiaJ compression in 
the beam, and if they are not applied at the center of gravity of the sec- 
tion they also produce bending about one or both axes of the cross- 
section, depending on the nature of their eccentricities. The calcula- 

22 



Art 24 Column Formulas 

tion of the moments and reactions for the beam and loads in Fig. 24 
will not be given. It should be noted that the right reaction has three 
components and the left, two components. The eccentricity of the 1000 
lb. load causes a moment about the Y—Y axis. The stresses on section 
A — A are given below. 

m x . x = 1358 in. lbs. 

niy.y = 500 in. lbs. 

P=1087 lbs. 

1358 X2.5 

f x . x = =163.0 

20.8 

500X1.0 
f y _ y = = 150.0 

3.33 

1087 

P/A = =108.7 

10 

Maximum Compressive Stress = 421.7 lbs. per sq. in. 
Maximum Tensile Stress = 313.0 — 108.7 = 204.3 lbs. per sq. in. 

At point a the maximum compressive stress occurs, and at point b 
the maximum tensile stress. 

The above procedure is satisfactory when the beam is relatively short 
so that the deflection of the beam is small. In the case of long, slender 
beams carrying a large axial load, additional moments are caused by the 
deflection of the beam, which may be considered as a laterally loaded 
column. This case will be treated in a later article on wing spars. 

24. Column Formulas — When a column is straight, and the only 
load imposed upon it is axial and is applied at the center of gravity of 
the column, then the stress equals P/A, where P is the load, and A the 
cross-sectional area. For very short columns the maximum value of 
P/A closely equals the stress at the yield point of the material, but 
for long columns the P/A at which failure will occur is much less than 
this value. The purpose of column formulas is to determine the max- 
imum allowable P/A. In practically all formulas the reduction in the 
ultimate stress for "column action" is a function of the ratio of the length 
of the column to its least radius of gyration. 

For columns having a high value of L/p, the maximum P/A is given 
very closely by Euler's formula. 

CttE cttEI 

P/A = or P = 



(L/p) 2 V 

E = Modulus of elasticity of material. 
L = Length of column. 
I = Least moment of inertia of column. 
p = Least radius of gyration of column. 

23 



Strength of Materials Art. 25 

c = A constant depending on the degree of fixity of the ends of 

the column, 
c = 1 for round ends. 

= 1/4 for one end free and one end fixed. 
= 2.05 for one end round and one end fixed. 
= 4 for fixed ends. 
The lower value of L/p to which this formula applies is given in the 
discussion below. 

The column formula that agrees best with test data on columns, 
whose L/p is less than the lower limit to which the Euler formula is ap- 
plicable, is J. B. Johnson's parabolic formula, 
f 2 

P/A = f (L/p) 2 

4ctt 2 E 
f = Yield point of material 

c = A constant with same values as in Euler's formula. 
For a value of L/p — 0, P/A equals the stress at the yield point, 
which for columns, coincides with the ultimate strength. As shown in 
Fig. 25, which is the curve for mild steel, the parabola is tangent to the 
Euler curve at approximately the point where columns cease to follow 
the Euler law. The abscissa of this point of tangency is given by the 



V 2 



expression L/p = \ ; the ordinate, for all degrees of fixitv, 

/ 

equals f/2. 

For mild steel, the assumption that the ultimate strength of the col- 
umn equals the yield point is perhaps somewhat severe. The use of 
40,000 instead of 36,000 for the formula in Fig. 25 would give less con- 
servative and yet very reasonable values. One of the important fea- 
tures of this formula is that the correction for column action is a func- 
tion of both the yield point and the modulus of elasticity. This is espe- 
cially valuable where experiment has not determined the column 
strength of the material, as in the case of high strength, alloy steels and 
duraluminum. 

25. Wood and Steel in Combination — There are occasional cases of 
wood and steel in combination when it is desirable to determine the pro- 
portion of the stress which each material carries. 

E w = modulus of elasticity of wood 

E s = modulus of elasticity of steel 

A w = area of wood 

A s = area of steel 

I w = moment of inertia of wood 

I s = moment of inertia of steel 
Case I. Beams subjected to bending. 

W = total load = W w + W s 

W w = load carried by wood 

W s = load carried by steel 

24 



Art. 25 



Wood and Steel in Combination 



w 



w 



W w = 



Wc 



E.I, 



1 + 



F T 
E.I, 



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Wp 

Fig. 25. 

Case II. Short columns 

P = total axial load 
P w = load carried by wood 
P s = load carried by steel 
P 

Pw = ; P s = 

E s A s 

l+ 

F A 

Case III. Euler columns 



/so 



eoo 



?<?o 



E w A, 



l + 



E,A. 



25 



Strength of Materials Art. 26 



p of composite column = 



4 / E w I w -j i !, 



E W A W + E S A S 

Total EI = (E W I W + E S I S ) 

26. Torsion, and Combined Torsion and Bending — In control sys- 
tems and surfaces tubes are frequently subjected to torsion. The stress- 
es due to torsion, and the torsional angle, or angle of twist for round 
tubes are given by the following formulas: 

f s = torsional unit stress 
J = polar moment of inertia of section 
M = twisting moment 
r = radius of the tube 

cf) = torsional angle in radians (this angle in degrees = 57.3 X<£)- 
L = length of tube for which torsional angle is desired. 
E s = shearing modulus of elasticity 
M.r M.L 

f s = and cf) = 

J E S .J 

The polar moment of inertia equals the sum of the moments of 
inertia of the section about any two axes at right angles to each other, 
or, for circular sections, twice the ordinary moment of inertia. 

These formulas are for solid or hollow circular sections only. For 
elliptical and rectangular sections the distribution of stress is different. 
With both types of sections the maximum stress occurs on the broad 
side of the section. In a rectangle the stress is a maximum at the center 
of each face and decreases to zero at the corners. The following formulas 
give the maximum stresses and angles of twist. 

b = short side of rectangle, or minor axis of ellipse, 
h = long side of rectangle, or major axis of ellipse. 
f s = maximum stress on section. 

Rectangle Square Ellipse 

4.5 M 4.5 M 16M 

fa = f s = U = 

b 2 h b 8 7rb 2 h 

The stress at the center of the short side can be computed from / s 
by multiplying f s by the ratio b/h. The angle of twist for a square sec- 
tion equals <f> = 7.1 1 ML/E s b 4 . 

Frequently a member is subjected to both bending and torsional 
stresses. These two kinds of stresses may be combined as follows: 

M-y 

f b = the bending stress computed from 

I 

M.r 
f s = the torsional stress computed from 

J 

26 



Art. 26 Torsion and Combined Torsion and Bending 



-±+Vv 



F c = f- \ f s 2 + fb 2 /4 = maximum tensile or compressive 

2 stress. 



F S = V* 



f b 2 /4 = maximum shearing stress. 



In using all the formulas given in this chapter the greatest care 
must be taken to have the units consistent. It is best to use the same 
units in all computations. A large part of the trouble with units is 
caused by the use of feet and foot units in places of inches and inch 
units. 

References: 

"Applied Mechanics,'' Vol. I and II, by Fuller & Johnson. 
"Theory of Structures," by C. M. Spofford. 

"Modern Framed Structures," by Johnson, Bryan & Turneaure. 
"Mechanics of Materials,'' bv M. Merriman. 



27 



CHAPTER II 

GENERAL CONSIDERATIONS 

27. Visibility — Visibility is a quality which varies in importance with 
the type of airplane and the service which it is intended to perform. 
The pilot must always have a field of view which will permit him 
to fly his airplane and land it with ease and safety. When the crew 
consists of three or more persons, a judicious arrangement of the per- 
sonnel will relieve the pilot of all duties except that of flying the airplane, 
and a careful arrangement of the remaining members of the crew should 
leave no blind spots which would permit an enemy aircraft to approach 
unseen. As the number of the crew is reduced, the importance of a wide 
field of vision for each individual member is increased, and in a single- 
seater pursuit, where the pilot must alscgassume the duties of observer, 
visibility attains its maximum importance and can be secured only by 
giving the greatest care and attention to the design of the structure as a 
whole. 

If the pilot has considerable freedom of movement, he may, by 
changing his position, look around almost any obstruction and reduce 
the blind spots to a minimum. Ordinarily, however, he is strapped in 
his seat with a belt so that his movements are restricted to bending at 
the waist and neck, and turning the head. The cockpit should be de- 
signed to enable the pilot to take the maximum advantage of such move- 
ments as are possible to improve his field of view. The cockpit open- 
ing must give ample room for movements of the head and shoulders. It 
should be cut out flush with the sides of a fuselage of rectangular cross 
section, and where the section is round or oval, the opening should be 
large enough to permit the pilot to look down vertically over the side. 
With monocoque construction it may prove impractical to cut out a large 
enough opening without seriously weakening the structure, but with a 
stick and wire or plywood truss fuselage, the cowling can be cut away as 
far as the top longeron without any difficulty. From a properly de- 
signed cockpit, the pilot should be able to look over the side and see 
directly beneath the fuselage. 

A narrow fuselage, especially at the engine section of a tractor air- 
plane, is favorable to a good field of view. It is difficult to secure good 
vision about the wide blunt nose of a rotary motor housing, while the 
narrow section possible where the cylinders of the engine are vertical and 
in line creates a very good field of view forward. A large nose radiator 
increases the blind area in front, the most favorable condition being 
reached when the engine cowling can be run down to a rough conical 
shape with the point at the hub of the propeller. While it is desirable 
to have an engine housing which will protect the engine from the weather 
as much as possible, the visibility can be greatly improved by reducing 
the volume enclosed by the cowling to a minimum. The cylinder heads 
may be left exposed, and along the sides of a vertical motor the cowling 
may be "dished in" to improve the field of view. 

To secure good visibility about a wing, it is essential that the pilot be 
so placed that his eyes are nearly on the line of the chord. If the trail- 

28 



Art. 28 Choice of Wing Section 

ing edge of the center section be cut away to enable the pilot to be 
properly placed, he can uncover all the blind area obstructed by the 
wing unless it is of extremely thick section. Perfect visibility can be 
secured with a parasol monoplane, or about the upper and middle wings 
of a biplane and triplane respectively. The narrow chord of the wings 
of a triplane tends to reduce the blind area obstructed by them to a 
minimum. Good visibility around the lower wing of a biplane is more 
difficult to secure. It is impossible to eliminate this blind spot entirely, 
but it may be cut to a minimum by careful design. The use of a nar- 
rower chord on the lower wing than on the upper, a wide gap which 
will place the pilot high above the lower wing, a smaller span to the 
lower than to the upper wing, and the cutting away of the trailing edge 
near the fuselage are devices at the disposal of the designer which will 
cut the blind area to a minimum. The use of a tapering wing with wide 
chord at the fuselage and the use of large dihedral in the lower wing are 
both factors which tend to increase the blind area under the wing. 

Stagger is an arrangement which greatly assists the designer in se- 
curing visibility by enabling him to locate the wings in the most advan- 
tageous position with relation to the pilot. A slight positive stagger is 
generally necessary if the pilot of a single-seater biplane is to be placed 
with his eyes in line with the chord of the upper wing. By use of stagger 
the lower wing may be placed so that the area obstructed by it is in the 
least important part of the field of view. A maximum angle of vision 
over the leading edge of the lower wing is, in general, more important 
than a large angle over the trailing edge, and an adjustment of the stag- 
ger may be used to place the lower wing in the position which gives the 
best results. 

External obstructions such as struts, gun installations, exhaust mani- 
folds, radiators, etc., interfere with the pilot's view to a greater or less 
extent. Where the obstructing surface is small and detached from larger 
obstructing surfaces, as is the case with narrow struts, a very slight 
movement of the pilot's head will enable him to see all the area behind 
them so that their effect on the field of view is unimportant. But in the 
case of radiators, gun installations and exhaust manifolds which are di- 
rectly attached to a large obstructing surface, the field of view may be 
materially affected. The use of short exhaust stacks and the housing of 
the guns inside the fuselage are desirable, while the building of the radi- 
ator into the wing so that it conforms to the wing curve undoubtedly 
gives the best installation from the visibility standpoint. 

28. Choice of Wing Section — There are three fundamental consid- 
erations in the aerodynamical design of an airplane: 

I. The determination of the wing section and area, and the dis- 
position of area which will give the desired performance in 
steady motion. 

II. The disposition of auxiliary surface area which will give 
proper stability and controlability. 

III. The reduction of parasite resistance to a minimum. 

The first problem is the chief concern of structural analysis. The 

29 



General Considerations 



Art. 28 



selection of the wing area and section is necessary for a determination 
of the span, chord, strut locations and spar sizes. . 

For a preliminary determination of the wing section, it is sufficient 
to consider various sections applied in turn to a single airplane of con- 
stant weight, area, power and parasite drag. Certain criteria may thus 
be developed which permit evaluation of the properties of the sections 
for high speed, climb, ceiling, etc. 

The assumption of constant area is a first approximation. For final 
choice the area must be considered as variable, and selection made by 
means of a series of performance estimates. However, the analytical 
tests used are adequate to determine those sections best adapted to each 
class of airplane. 

For level flight at high speed 

P = S(K x Vc)AV 3 
where P is the power of the motor. 

K x is the drag coefficient of the wings. 
K y is the lift coefficient for the wings. 
K ym is the maximum value of K y . 

is the parasite coefficient per unit wing area. 



or 



is the wing area, 
is the air density, 
is the speed — in thi 

(K x + c) 
8K V A V 3 



case high speed. 



f P 1 % f ^ 

and V= J 1 I 

lK y 8AJ [K x +c 

It is evident from this equation that if these airplanes, differing only 
in aerofoil section, be flying at the same value of K y , the airplane with 
aerofoil of higher K Y /K X at that condition will not need all its power to 
maintain the speed, or it may be driven at a higher speed for the same 
power. A preliminary valuation of aerofoils for high speed is the value 
of K y /K x at the K y reached in flight. 

The landing speeds of two airplanes differing only in aerofoil section 
are inversely proportional to the square roots of the maximum values 
of the lift coefficients of the sections. 



V 



w 



8K y] 
W 



1 p 



J A J [K ym J 

The maximum unit lift has been listed, therefore, in Table I. 

Speed range is defined as the ratio of high speed to landing speed. 
Since landing speed is inversely proportional to the square root of the lift 

30 



Art. 28 



Choice of Wing Sections 



coefficient, and high speed to the cube root of the L/D at the value of 
K y usual to high speed, the speed range is proportional to the product of 
these figures: 

r i i % 

V m o= 1 

Voc (L/D) 1 - 
V/V m o E (K ym )^X (L/D)* 

The values of this quantity are listed in Table I, for K y = .0008 and 
.0004. It will be observed that the aerofoils are not of the same value 
in speed range for low speed and high speed airplanes. 

Due to the fact that this speed range factor is a first approximation, 
one aerofoil should not be chosen above another by reason of a small 
favorable difference, for finer analysis may reverse the result. However, 
large differences shown by this method may reasonably be taken as in- 
dicative of comparative merits. 

The maximum altitude or minimum density which an airplane can 
reach is a complex function of its engine power and drag characteristics. 
In his paper on "Supercharging," Dr. de Bothezat shows that the ceiling 
density is expressed 



WoY 



P % A* *% 



X 



fKx + cl- 



where n is the average propeller efficiency. 

In the comparison being made, the airplane with the aerofoil possess- 

IV 2 

ing highest value of will give the highest* ceiling. This quantitv is 

K x 
given in Table I. 

The rate of climb is also a complex function of the power and drag 
characteristics. 



nP r 



R 



W 



[£)•* mi 



As long as the power and wing area are considered constant the rate of 
climb may be taken as dependent solely on the quantity, K x /K y 3/2 and 
in the first approximation is maximum when K 7 * /2 /K x is a maximum. 
This quantity is, therefore, also a criterion of climb. 

In an analysis of the cruising problem, Lt. Col. Durand shows that 

the maximum radius is obtained at an ande of attack of min- 



So Ions: as c is actual 



constant the aerofoil which has the 
31 



General Considerations Art. 28 



highest maximum K y /K x will give best cruising radius. L/D maximum 
is, therefore, listed as a criterion for cruising radius. 

A minimum c.p. movement on the aerofoil is desired in order that 
the load may be at all times well distributed between the spars. A large 
movement may be a sufficient cause for the rejection of a section even if 
its performance characteristics are satisfactory. The c.p. travel from the 
most forward position to K y , .0008, has been computed for the aerofoils 
listed. This is not the range applicable to all types of airplanes, but 
gives a good comparative figure. 

Spar depth is a very important consideration. It is possible to secure 
very desirable aerodynamical characteristics with very thin or very 
highly cambered sections which would be entirely inapplicable to any 
but the most lightly loaded airplanes. For most conventional two-spar 
designs the usual spar locations are at about 10 per cent to 15 per cent 
and 70 per cent of the chord. The spar depths of the sections have 
therefore been listed in Table I. Other things being nearly equal, the 
section with the greatest spar depth is to be recommended. 

it is rarely the case that an aerofoil is selected for a single per- 
formance characteristic. Few airplanes require predominantly high 
speed, climb, or ceiling, but rather varying combinations of these qual- 
ities. 

In Table II the fifteen Air Service Specification types have been di- 
vided into six general classes with the average performance require- 
ments shown. 

It is obviously impossible for one rigid aerofoil to be best "all 
around." Only by variable area and camber arrangement could such 
a condition be more nearly fulfilled. The choice must be made rather as 
a compromise of performance characteristics, with requisite spar depth 
and reasonable c.p. movement. 

On the basis of the above figures the sections given in Table II are 
recommended as suitable for the six classes. It will be observed that in 
some cases two sections are listed, which indicates that their character- 
istics are very similar and little choice is possible. The complete char- 
acteristics and ordinates for these sections are given in Table III. 

In a pursuit airplane speed must be combined with fast climb and 
high ceiling. The wing section must, therefore, be somewhat of a com- 
promise between an extremely high speed and high climb wing. There 
are, however, different types of pursuit airplanes and the relative im- 
portance of speed, climb and maneuverability varies with the special 
service for which the airplane is intended. Where ability to maneuver 
is to be strongly emphasized, a section possessing high lift characteristics 
is employed. The U.S.A.-27 is well adapted to this use, and the depth 
afforded for spars makes it suitable for internally braced construction. 
This section is also very efficient except for extreme high speed. It is 
well suited to a night pursuit airplane in which high speed is not so im- 
portant as climb, low landing speed and maneuverability. For a pur- 
suit airplane that would be required to land in restricted space, as in 
mountainous regions or roads, it is essential to use as high a lift section 

32 



Art 28 Choice of Wing Sections 

as possible in order to reduce the area and hence the span. For this 
purpose, also, the U.S.A.-27 is the best wing. The U.S.A.-16 has better 
climb and ceiling characteristics than the R.A.F.-15, as is indicated by a 
higher value of K y s/2 /K x , and is therefore to be chosen if more im- 
portance is to be attached to climb and ceiling than to high speed. The 
U.S.A.T.S.-5 is too thick for a pursuit airplane, but is mentioned to 
indicate the best that can be done in adapting thick sections to speed 
work. The same sections are adapted to class 2, because the same re- 
quirements apply except to a lesser degree. 

For both classes 3 and 4 the U.S.A.-5 was selected as the best thin 
section. It combines low speed with good climb and high efficiency at 
cruising speed, characteristics which are of more importance than high 
speed. Since a low landing speed is very desirable with airplanes of 
these types, and since also it is important to keep the. span and hence 
the area as small as possible, the large Ky of the U.S.A.-5 section is one 
of the most important factors to be considered. However, for airplanes 
in which high speed is of more importance than the considerations dis- 
cussed above, the U.S.A.-15 section is to be recommended. Its structural 
characteristics are superior to those of the U.S.A.-5. In these types of 
airplanes it is very desirable to use a single instead of a double bay cel- 
lule, and the greater spar depth of the U.S.A.-15 section may make this 
possible in cases where, with the U.S.A.-5 section, it would be imprac- 
ticable. 

It will be observed that the only intermediate and thick sections 
chosen for any class were the U.S.A.-27 and the U.S.A.T.S.-5. An ex- 
amination of Table II will show that these are far superior to the 
other sections in the groups of intermediate and thick wings, for all 
except very special work. For small or moderate sized airplanes 
these wings will be too deep to be economical unless at least partially 
internal braced construction is used. 

In the larger designs of classes 5 and 6 advantage could be taken of 
the greater depths of these sections to increase the length of the bays, 
thereby reducing the inter-plane bracing. In all types of airplanes, but 
particularly in larger bombers and express or passenger airplanes where 
the value of the airplane is almost directly proportional to the useful load 
carried,. the advisability of selecting a high lift section becomes apparent. 
For instance, the U.S.A.-27 has a K ym 42 per cent greater than the 
U.S.A.-16, and therefore the same wing area can support a proportion- 
ately larger load. But, as the total useful load for an airplane is rarely 
as much as 40 per cent of the gross weight of the airplane, the use of a 
high lift wing would about double its net carrying capacity. This is, of 
course, partly offset by the fact that the ceiling and the climb, which are 
closely proportional to the power loading, would be considerably af- 
fected by increasing the gross weight of the airplane. In commercial 
airplanes this is not as important as in military types. The change in 
high speed, moreover, would not be great. In commercial airplanes and 
in long range bombers where economy in operation and large radius of 

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36 



Art. 28 Choice of Wing Sections 

action are more important than high speed, the maximum L/D is the 
chief criterion, provided the c.p. travel and the spar depth are reasonable. 

For bombers, both U.S.A.-5 and 15 are recommended. The greater 
maximum lift coefficient of the U.S.A.-5 permits the wing area to be 9.5 
per cent less than with the U.S.A.-15 section, with the same landing 
speed and nearly the same ceiling. In the Bothezat formula for ceiling 
density it should be noted that the ceiling is closely proportional to the 
cube root of the area and to the two-thirds power of K y s/2 /K x . There- 
fore, within comparatively narrow limits this formula may be used to 
compare the ceilings given by different wings when the wing areas are 
adjusted so as to give the same landing speed in each case. Since the 
efficiency at cruising speed, and the speed range of these two wings are 
nearly the same, the U.S.A.-5 section is recommended as a good alternate 
to the U.S.A.-15, providing the structural proportions of the airplane 
are such that the smaller spar depth afforded by the U.S.A.-5 is suf- 
ficient. 

There are also types of airplanes which are designed for special pur- 
poses and a general purpose wing is not required. In a racing airplane 
the value of L/D at A" y , .00035, is almost the sole measure of the suit- 
ability of the wing. In an airplane intended for very high ceiling, max- 
imum value of K/ n /K x is the important factor. For an "all purpose" 
airplane of moderate size the U.S.A.-15 is recommended as the best 
"thin" section from both an aerodynamical and structural point of view. 

The disposition of wing area (whether a monoplane, biplane or tri- 
plane is used, and the aspect ratio, gap, stagger, etc.) has a definite effect 
in modifying the L/D given by model tests. Quantitative studies of 
these combinations lead to the following conclusions: 

I. The aspect ratio of the wing should be as great as is con- 
sistant with structural strength, visibility and other require- 
ments. With increasing aspect ratio: 

(a) The lift improves steadily for all of the flying angles. 

(b) Maximum lift occurs at a smaller angle. 

(c) L/D improves steadily. 

II. The gap-chord ratio should be as large as possible. With in- 
creasing gap-chord: 

(a) Lift improves steadily. 

(b) L/D improves, especially at low angles. 

III. As large a positive stagger as is consistent with structural 
and balance requirements should be used. For by increase 
of stagger: 

(a) Lift improves. 

(b) L/D improves at low angles. 

(c ) c.p. movement is decreased and made more nearly 
stable. 

37 



'.A 



TABLE III 

U.S.A. 5 

M.I.T., March 17, 1917 

Velocity— 30 M.P.H. in all tests 

All Models— 18" x 3" 





Characteristics 




Ordinates 












Distance 


Lower 


Upper 


i 


K y 


L/D 


C.P. 


from 


Ordi- 


Ordi- 










L.E. 


nate 


nate 


—4 


— .000326 


—1.58 


0.00 


.62 


.62 


—2 


.000346 


3.64 


.753 


1.25 


.13 


2.10 





.000910 


12.28 


.498 


2.50 


.03 


3.03 


2 


.001355 


15.72 


.415 


3.75 


.00 




4 


.001740 


15.98 


.348 


5.0 


.03 


4.40 


6 


.002120 


14.80 


.327 


7.5 


.25 


5.40 


8 


.002470 


13.52 


.315 


10. 


.57 


6.20 


10 


.002870 


12.08 


.303 


20. 


1.55 


7.92 


12 


.003130 


10.84 


.300 


30. 


2.02 


8.30 


14 


.003285 


9.25 


.288 


40. 


2.17 


8.14 


16 


.003205 


7.63 


.298 


50. 


1.96 


7.55 


18 


.003150 


4.57 
M.I.T. ; 


.330 

U. S. A. 15 
, November 


60. 

70. 

80. 

90. 
100. 

Radius 
Radius 

, 1917 


1.56 
1.16 

.76 

.55 

.25 
L.E. = 
T.E.= 


6.75 
5.63 
4.24 

2.52 
.25 
.36 
.25 


—4 


— .00035 


—2.8 




0.00 


1.06 


1.06 


—2 


.00008 


1.4 




1.25 


.46 


2.56 





.00058 


9.97 


.480 


2.50 


.33 


3.44 


2 


.00108 


15.50 


.407 


5.0 


.15 


4.80 


4 


.00147 


16.50 


.353 


7.5 


.03 


5.58 


6 


.00183 


15.25 


.330 


10. 


.00 


6.11 


8 


.00220 


13.95 


.320 


15. 


.12 


6.88 


10 


.00257 


12.35 


.310 


20. 


.33 


7.28 


12 


.00290 


11.30 


.303 


30. 


.86 


7.61 


14 


.00301 


9.43 


.297 


40. 


1.00 


7.55 


16 


.00286 


4.98 


.340 


50. 

60. 

70. 

80. 

90. 
100. 

Radius 
Radius 


.72 

.28 

.00 

.11 

.23 

.50 
L.E.= 
T.E.= 


7.11 
6.36 
5.32 
3.90 
2.50 
.83 

.51 

.20 



38 



TABLE III {Continued) 

U.S.A. 16 
M.I.T., October 15, 1917 

Characteristics Ordinates 











Distance 


Lower 


Upper 


i 


K y 


L/D 


C.P. 


from 


Ordi- 


Ordi- 








.61 


L.E. 


nate 


nate 


_2 


— .00003 


—0.5 


0.0 


.29 


.29 





.00035 


7.5 


.39 


2.5 


— .64 


2.99 


2 


.00092 


18.9 


.335 


5.0 


— .76 


3.88 


4 


.00135 


16.7 


.315 


7.5 


— .83 


4.52 


6 


.00170 


15.0 


.300 


10. 


— .83 


4.95 


8 


.00209 


13.7 


.290 


15. 


— .76 


5.42 


10 


.00241 


11.6 


.285 


20. 


— .51 


5.67 


12 


.00252 


7.0 


.300 


30. 


— .06 


5.86 


14 


.00247 


4.7 


.330 


40. 


.00 


5.74 


16 


.00234 


3 .5 


.37 


50. 
60. 
70. 
80. 
90. 
100. 


— .19 

— .70 

— .76 

— .70 

— .45 
.43 


5.29 
4.65 
4.01 
3.25 
2.23 
.43 










Radius L.E. = 


.57 










Radius T.E. = 


.43 



U.S.A. 27 
M.I.T., March 3, 1 ( '20 



—6 
—4 
—2 

2 
4 
6 
8 

10 
12 
14 
16 



00047 
00012 
00062 
00099 
00140 
00179 
00213 
00251 
00284 
00319 
00346 
00359 
00350 



-J. 3 

1.1 

8.0 

13.6 

16.1 

15.7 

14.6 

13.4 

12.0 

10.8 

9.9 

8.6 

6.8 



56.5 
43.3 
38.3 
36.0 
34.7 
34.0 
33.3 
33.0 
33.3 
33.7 
34.0 



0.00 

1.25 

2.50 

5.0 

7.5 

10. 

15. 

20. 

30. 

40. 

50. 

60. 

70. 

77. 

80. 

90. 
100. 
Rad 
Rad 



1.77 

.50 

.33 

.17 

.10 

0.00 

.10 

.35 

.95 

1.17 

.80 

.25 

.10 

0.00 

.05 

.15 

.67 

ius L.E. 

ius T.E. 



1.77 

3.80 

5.07 

6.93 

8.22 

9.17 

10.50 

11.33 

11.90 

11.57 

10.77 

9.52 

8.00 

6.03 
3.65 
.67 
1.06 
.21 



39 







TABLE III {Continued) 










U.S.A.T.S. 5 












M.I.T, 


, February, 


1919 








Characteristics 






Ordinates 












Distance Lower 


Upper 


i 


k y 


L/D 


C.P. 


from 


Ordi- 


Ordi- 






1.8 


■ 


L.E. 


nate 


nate 


—8 


.00017 


0.0 


2.50 


2.50 


—6 


.00051 


5.8 


.91 


2.5 


— .80 


5.69 


—4 


.00088 


9.6 


.62 


5.0 


—1.80 


7.50 


—2 


.00126 


11.8 


.50 


10. 


—3.00 


10.10 





.00164 


12.6 


.44 


15. 


—3.40 


12.00 


2 


.00200 


12.8 


.40 


20. 


—3.50 


13.30 


4 


.00237 


12.2 


.38 


30. 


—2.90 


14.60 


6 


.00271 


11.3 


.36 


40. 


—1.50 


14.68 


8 


.00303 


10.5 


.355 


50. 


— .60 


13.80 


10 


.00336 


9.9 


.35 


60. 


— .40 


12.40 


12 


.00362 


9.3 


.34 


70. 


— .20 


10.40 


14 


.00385 


8.5 


.33 


80. 


.00 


7.76 


16 


.00379 


7.3 


.32 


90. 
100. 


.00 
1.00 


5.00 
1.00 










Radius L.E. = 


3.00 










Rad 


ius T.E. == 


= 1.00 



R.AJ. 15 
M.I.T., December 30, 1919 



-2 

2 
4 
6 
8 

10 
12 
14 



00004 
00033 
00085 
00122 
00160 
00194 
00228 
00260 
00240 



8.1 
16.6 
16.5 
15.4 
13.4 
11.9 
10.7 

5.0 



443 
367 
327 
310 
300 
297 
290 
350 



0.00 

1.25 

2.5 

5.0 

7.5 

10. 

15. 

20. 

30. 

40. 

50. 

60. 

70. 

80. 

90. 

95. 
100. 



.27 

— .47 

— .70 

— .93 
—1.00 
—1.00 

— .78 

— .50 

— .06 

— .03 

— .20 

— .50 

— .67 

— .58 

— .35 

— .16 
.42 



Radius L.E. 
Radius T.E. 



.27 
1.86 
2.78 
3.95 
4.62 
5.05 
5.53 
5.75 
5.83 
5.63 
5.30 
4.80 
4.13 
3.26 
2.23 
1.74 
.42 

.45 

.42 



40 



Art. 29 Characteristics of Military Airplanes 

29. Characteristics of Military Airplanes — In Table IV are given 
data on most of the best airplanes of various types of both American 
and foreign design. The main purpose of the table is to present the 
structural characteristics and loadings of these airplanes and show the 
performance obtained. The horsepower loading is somewhat mis- 
leading, except in comparing airplanes with the same type of engine, 
because the loading is based on the rated power of the engines rather 
than their maximum, and the margin between the rated and maximum 
horsepowers varies considerably with different engines. Furthermore, 
even with a particular engine, the type of propeller that is used makes 
a very appreciable difference in the performance of the airplane. It is 
only by eliminating differences in their power plants through the use 
of a similar propeller and engine that the performance of two airplanes 
can be fairly compared. In columns 7-9 where two values are given 
the second value is for the lower wing. 

30. Structural Weight Analysis of Airplanes — Table V is a tabula- 
tion of weights for several different types of airplanes which will aid in 
preparing preliminary weight estimates for new designs. The total 
weight is divided into several main groups, such as power plant, fuel 
and oil, etc., and each group is expressed as a percentage of the total 
weight. 

P = Total weight in lbs. of engine or engines, including carburetors, 
ignition system, self-starter, radiator and connections, water, 
propeller and propeller hub. 

F = Weight of the capacity of gasoline plus oil. 

I' = Weight of crew and all movable equipment such as bombs and 
racks, ammunition, baggage, guns, wireless, sights ,and camera. 

B = Weight of fuselage complete, with all it contains except P, F, 
and U. This weight includes structure, doped covering, cowl- 
ing, tanks, engine mounting, seats, controls, instruments, 
gun mounts, tail skid, etc. In case of multi-engined airplanes, 
it includes weight of cowling and tanks. 

W = Total weight of main wings, including ailerons, interplane 
struts, wiring, fittings, center panel and center panel bracing, 
drag bracing, tip skids, etc. 

C = Total weight of landing gear — everything beyond fuselage and 
wings. 

E = Total weight of empennage, including bracing exterior to the 
fuselage. 

31. Load on Wings and Weight Per Horsepower — The loadings for 
different types of airplanes of American, English, French, Italian and 
German design are given in Table IV and the table on page 44 and may 
serve as an approximate guide in the selection of the wing area or of 
the size of engine to use. 

41 



General Considerations 



Art. 29 







IA6U -J CRARA3I2SIST 


I0S or H1UTART AJEPLAKES 




















lame at Airplane * SBgiM 


?ype 


Penctlen 


•eight 


Ueeful Sreaa 


3p«n 


Chord 


lapect 


Oap 


Eorae- 


Lb../ 


Lb../ 


jrouad 


Jerri oe 










3ipty ♦ 


Loaa rtleht 






BUI. 










^peed 


Cilia,, 


,-s 


1 


2 


S 


4 


,, b 6 


7 


8 




10, 


xl 


12 


13 








Thomaj-korM.Eiapano "ff« 


as-s 


1-Fnrenit 


1606 


690 2096 


26.0 


5.25 


4.96 


4.68 


300 


8.3 


7.0 


152 


23,700 




*OrdBSIlM,Kl»pSBO "H" 


D 




1776 




30.0 


6.00 


6.00 


4.33 


SCO 


J. 3 


6.1 


147 


22,000 


3 


Verrllle.Hlapaao "H" 


7CF-1 


1-Por.alt 


1980 


640 2*20 


32.0 


4.76 


6.75 


4.5 


300 


9.7 


6.75 


150 


22,40 >. 


* 


Loeniag monoplane. 


IU 


1-Pur«ult 


1466 


976 2640 


32.76 


7.00 


4.66 




300 


12.3 


6.9 


143.5 


18,500 


6 


Poxrllo.Llberty "8" 


PTL-8 




1716 


660 2286 


26.67 




4.9 


5.25 




8.06 


7.9 








S.I.5A (i»r.) Siarino -B" 




l-Pareult 


jjU 


676 2060 


ii.-'i 


6.0 


4J3 


4.63 


180 


S.4 


11,4. 




01 ,v: 


7 


Sped Herbemont.Biepano 


200-2 


2-Pnrenit 


1866 


1015 2880 


51.9 1 
26.41) 


5.74) 


5.651 


6.10 








— 142 — 












186i 


790 2*65 




5^6} 




8.2 


8.86 




28|2CC-" 




Spad, Hiepano 8-BEC 


150-1 






660 18L6 


26.26 


5 '92 


6.S5 




220 




8.26 




9 


Me?port,Oname MoaOTal-re 


29 


1-Pur«ult 


9*0 


676 1636 


24.5 




6.25 


5.92 




8.9 


1C.2 


















25.5 


3.28 










10 


•Kieuport.Hlapano 8-?B 


29C-] 


1-Pnrrnlt 


1480 


746 2426 


31.83 








300 




6.1 




24,900 


11 


Spsd,Hlepaao 5-xs 


70-1 


1-Poxeolt 


1100 


460 1560 


25.75 
25.0 


6.17 
4.6$ 


5.C 
6.45 


3.56 


200 


8.0 


7.2 


124 


21,000 


12 


K-oj-dot uoaorclane 
HI apeno T 


C-l 


1-Par^t 


1440 


410 I860 


29.53 


S.17 


4.6 




180 


10.3 


10.2 


162 




If 


i.oraoe Xoaoii. Jono.. icom. 


lie. 


1-pnjtnlt 




60C 1425 


27.76 


5.33 


6.2 






9.8 


9.6 


140 


23,000 


14 


S.7.A.,le0tt» PTaeoalni 7-6 




l-Furenit 


1690 


720 2310 


30.0 


6.06 


5.96 


5.86 


260 


8.8 


8.9 


147 




IS 


5.V-1.,- SJ>U. Blglae 




1-Pnre-uit 


1446 




24.21 


6.26 


6.C 




220 




8.95 






16 


AB.aldO,- SJ.A. Blglne 




1-Porealt 


1470 


496 19*6 


26.0 








220 


8.7 


6.9 


137 




17 


•sopeith, Hiepano 6Pb 


Dolphin 0-1 


1-Pnrenlt 


1616 


790 2306 


52.5 


4.60 


7J 


4.26 


300 


8.76 




139" 


24,600 




Sopeith.Clergat 




1-Pnrenit 


976 


510 14S5 


28.0 


4.60 


6.2 


4.50 


150 


6.46 


11.4 


119 


19,000 


19 


Sopwlth,BB>2 


summ*.* 


1-Armea-ed 
freneh Plgate 


1846 


670 2616 


50.67 


5.00 


..15 




200 


9.4 


12.6 


126 


13,000 


20 


Sopeith BB-2 


Snipe 7.P.1 


. 1-Purenlt 


1210 


786 19*6 








4.26 




7.2 


9.8 


124.5 


20.000 


21 


Martlneyde ?all» 3. yaleea 




L-Psreult 


1790 


636 2326 


31 .0 








220 


6.9 


10.6 


138" 


22,500 


22 


•i:l3=port,A3C Sedial 


Slgnt HAW*. 


1-ParejUt 


1546 


736 2100 


28.0 


5.25 


6.36 


4.5 


320 




6.6 


162 


29,000 


23 


pt-.i.,y.ro.d.. 


0.141 


l-purenit 


ISM 


626 2066 


30.83 
26.64 


5.44 
3.92 


6.7 
6.8 


4.67 


1*0 


8.66 


12.8 


112 




24 


? otter, e.rc.dee 


B-T 




16*0 


446 2006 


29.29 


6.25 


3.4 


4.53 


ISO 


8.1 


11.1 


117 
















22.92 


3.93 


6.76 












* 


•u.s.X. bl - a , .-.lupt.o -«- 




1-rigit 


1010 


9*5 2996 


39.33 


6.60 


7.U 


6.42 


300 


T.t 


10.0 


124 


20, 900 


M 


LePere, Liberty -12- 


O.S.A. 0-U 


2-fig*** " • 

ObaarrmtlOB 


26*0 


11*6 3T46 


41*! 


6.43 


7.4 


3.05 


400 


9.06 


9.36 


133 


20,200 




E.3.A. D-t 




2-OeeerreUen 


2590 


UK 36M 


42.68 


6.60 


7.76 


6.6 




8.16 


8.98 


126 


20,000 


28 


LePere Trlrlane. 2- 

Llbertr "12e" 


U.S.A. 0-11 


S-Ar>9- Carp. 
Obearratla 


6466 


S120 (5T6 


64.6 


6. 73 


9.6 


6.1T 


800 


9.98 


»o .7 


112 


16.800 


29 


5r«)aet, Llhrtj -12- 


14- At 


2-Day Boaiaer 

A Beoen. 


2390 


1330 3T70 


47.25 
40.47 


4.68 
t.25 


74 
6.6 


e J* 


400 


7.16 


9.4 


1*0 


21,800 


3D 


■Breqnet, Renault UK* 


17-ca 


2-Sagr Baeaer 


2*40 


1410 y„ 


44.86 








460 


8.6 


9.06 


137 


19,700 








A Euan. 






















SI 


Spad, Hiepano S-BZC 


11-A2 


2-oaMrretlon 


MM 


825 2310 


St.iT 


4.63 


8.0 




22C 


7.16 


104 


116 


21,000 


32 


Sal*.-, salxaon C.U.9I 


4-AB2 


T-ranoh ?lghtei 


3266 


1156 4410 


49.9 


6.3 


7.36 




230 


7.26 


19.1 


924 




S3 




2-Al 


2-Oaeerratlen 


1676 


1126 MOO 


38.6 


6.61 


7.3 


6.61 


2S0 


7.0 


12 4 


119 


20400 


St 


•iorane Seulnler. LI tarty "12 


22-C2 


2-obeerratlan 




1246 3906 










400 




9.76 


1344 


19,400 


56 


ta "" 1 " r " 4 " 


5.11T 


A Plater 


(440 


1000 3440 


41.5 
40.0 


5.«7 
4. S3 


7.3 

9.2 


3.00 


240 


9.46 


134 


110 




S6 


l.v.0.,8^ 


0-4 


2-0baer»ation 
4 Plghter 


2090 


950 3040 


42.76 
40. Tl 


6.26 
4.46 


8.16 
8.76 


4.T6 


220 


5.56S.13J 






ST 


•Halberetadt, Bern 


0-6 


2-ObaerTetlom 
A rlgnter 


20*P 


960 SC10 


42.08 


6.33 


7.9 


8.03 


220 




13.7 






SB 


Bal beret adt, uercedea 


CL-2 


2-Flghter 


1766 


815 2670 


36.27 
54.92 


6.27 
4.23 


J'.l 


iM 


130 


8.3 


14.3 






39 


0.3. D-92, Libert; -12" 




2-0aMrr*tI<B 
A Day Bomber 


2816 


1506 4S2C 
2065 4870 


46.92 


6.76 


8.0 


4.12 


400 


8.3 

9.9 


104 
12.2 


126 
121.6 


18,700 
14,400 


40 


Glen Martin,2-Llberty "L2i" 




S-Eay Bomber 




3626 10226 


71.42 


7.92 




343 


800 


945 


12.6 


106 


10,300 


♦I 


poclllo. Libertj -12- 


fTl^li , , ,-, 


2-Day sce*er 


2826 


1725 466C 


43.26 


6,03 


Ml 


..'.ft 


4CC 






111 


13.700^ 


42 


71okere visy, Larri Sunbeama 


TB-2Te 


S-Slght Bomber 


6736 


3666 10800 


6T.1T 


10 J 


6.4 


10.17 


6O0 


53" 


20.4 




8,000 


45 


•viciere Timy,2-Llberty "12e- 




3-Nlght Bomber 


6700 


64S0 12100 


4T.1T 


1C J 


*•* 


10 .IT 


SOC 


9.18 


16.1 




10,500 




Handley-Page,4 BJl^agla -8" 


7-1600 


7-ilsJit Bomberl6210 


•490 24T00 


L26rf 










8.56 


1T.1 


102 




.46 


Handloy-Pafie^-Llborty -L£j- 




4-BlKht Bomber 


6720 


6T06 J4426 


100.0 


.0,0 


10.0 


Urf 


300 


8.7 


18.0 


94 


T.400 




Caproal Biplane, 






























S-Llberty -12a- (3*Tyl 


OA-6 


4-»l«Jit Bomber 


7T00 


4*60 12260 


T4.TT 


9.12 


•at 


9.11 


1060 






108 




47 


Caproal Biplane. S-Plata 






























A-12 61a 


0a-6 


S-Hltfit Bomber 


"SO 


4400 USSO. 


76. TT 


Ft** 


, '.•■r. . 


I« U 


900 




12.6 


100 




48 


Caadron, 2-Salnieon C.C.9Z 


C-23 


Klght Bomber 


5C60 


4020 9070 


T7.0 
73.0 


• J.1 
4.71 


9.8 
U.1 = 


440 


3.4 


19.7 


96 




I" 


a. E.G., 2-Iieroedee 


* 


S-tl^rt Boirbe 


6390 


5080 »470 


40 .It 

51.37 


7.54 

-.a 


7.96 
T.l 


3.0 


820 


11.4 


144 


90 




'so 


Sothn, 2-¥eroeaee 


o-s 


4-Slgbt Barbel 


6040 


2720 37*0 


7l!78 
78 .0 


7.36 


10 J 
t.73 

.0r« 


7.0 


620 


8.9 


16.8 


73- • 




fl 


PrledrtohehATon. 2-Ueroedee 


2-3 


4-El(dit Bombej 


6930 


2716 8646 


7.47 


T.0 


620 


946 


14.4 






62 


*Vought, Hiapano "E" 


71-7 


2-Ai-rmoea 






























Training 


15*0 


636 2096 


54.11 


4.42 


7J6 


4.4T 


180 


7.36 


11.6 


114 


IT ,O0C 


6S 


CTdnsnoe, LeRhane 


C 


l-Advanoed 


836 


280 1116 


2C.0 


4.0 
S.T5 


4.3 

4.93 


3.83 


80 


6.36 


14.0 


98 


IS, 600 


M 


Thornae-Horae , Gnome 


3-4C 


1-Tralclag 


960 


396 1366 


£i.6i 


l.U 


4.33. 

4.26 


4.60 


100 


6.78 


134 


100 


14,800 
lAbaolate 


66 






1-Tralnlng 


(SO 


316 1146 


24.0 


3.60 


t.M 


4.00 


30 


T.6 


14.S 


100 


64 


CTartl... Hlapano -A- or "1" 


JI-6EB 
JB-4H 


2-T raining 


1796 

1S96 


890 24*6 

650 2146 


45.471 

34.T1] 


4.H 


;:J} 


6. IT 
5. IT 


180 


4.1 


1T.9 
14.3 


79 

93 


6,700 
8,000 




Cnrtlaa. Curtlee 0-X.5 


JB-4 D2 


2-Tralnlng 


1626 


490 2016 


43. 4T 


4.9* 


3.3 


8. IT 


90 


6.7 




73 


6,600 














34.71 


7.0 














6* 


*atto, leSnone 


504-K 


2-T raining 


1230 


600 USD 


34 .0 


4.45 


T.46 


340 


lie 


646 


It. 6 


90 




69 


Bienport, LeRhone 


24 


Trainlni; 


780 


426 1206 


2T.0 
26.42 


4.64 
2^3 


6.93 


3.92 


110 


7.6 


10.9 


103 


22,600 












* lacludee 


water 




























• fnaae airplane • 


an <rf 


arpMial 


aorlt. 






















•• speed « 


10,000 


t,«. 


























••• Stripped 


of all 




—rj eanlpmeBt. 













42 



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45 



General Considerations Art. 32 



No. of 

Ma- Type 

chines 



Weight per sq. ft. 
of Wing Area 


Weight 
Per H.P. 


Range 


Average 


Range Average 


6.2— 9.2 


8.3 


6.8—11.7 9.2 


6.9— 9.4 
7.5—11.6 


8.2 
9.4 


9.0—14.8 11.8 
11.4—18.7 15.9 



9 One-place Pursuit 

8 Two-place Pursuit and 

Reconnaissance 

6 Heavy Bombers 

32. Aspect Ratio, and Shape of Wing Tips — The aspect ratio of air- 
planes is affected by four main factors: considerations of visibility, 
structural weight, and aerodynamic efficiency and stability. In general, 
a high aspect ratio is favorable as regards visibility and aerodynamic 
efficiency and stability, but unfavorable from the structural point of 
view. The relative importance of these different factors varies with 
the type of airplane. In a commercial airplane, for instance, visibility 
in landing is even more important than in a military type, but visibility 
in other directions is not so essential. If a low aspect ratio is decided 
upon, care should be taken to select a wing with as small a center of 
pressure travel as possible, in order to secure good stability. In mono- 
plane construction, and particularly in internally braced wings, low 
aspect ratio is important, as it is in single bay airplanes which are close 
to the economical limit for single bay construction. With suitable wing 
tips the effect of aspect ratios greater than 5.0 on aerodynamic effi- 
ciency is slight. It will be noticed from Table V that the MB-3, one 
of the fastest airplanes built, has an aspect ratio of only 4.95. For an 
extreme low limit that might still give satisfactory results a value of 3.5 
is suggested. Another reason for a very low aspect ratio and conse- 
quently short span is the necessity, with certain types of both military 
and commercial airplanes, of landing in restricted areas such as moun- 
tainous regions or roads. Short spans also simplify the problem of 
housing an airplane. A method for increasing the effective aspect ratio 
of a wing of a given span and area is to taper it in plan form. It is not 
the average chord, but rather the chord at a distance from the tip equal 
to about .8 of the chord at that point, that determines the aspect ratio. 

Wind tunnel tests on the effect of the shape of wing tips show that 
the most efficient tip is in the form of a semi-ellipse. For the most part 
it is undesirable to have as long a tip as is necessitated by an ellipse, 
and a semi-circular shape is used to good effect. The chief disadvantage 
of either of these types is the difference in length between the front 
and rear spars. This makes it difficult to proportion the cantilever span 
and the inner bays economically. The rear spar will be weak at the 
outer strut point, and the front spar in the bay adjacent to the cantilever 
span. The wing with a square tip, just having the corners rounded off, 
remedies this rather serious defect. Another objection to the elliptical 
or circular tip is that with such a tip a considerable area is cut from an 
aileron just where it is most effective. One of the most efficient high 

44 



Art. 33 Location of Lift Wires in Inner Bay 

speed airplanes, the Curtiss 18-B, has this square tip slightly modified. 
As has been suggested, the influence of the shape of the tip is more pro- 
nounced with low aspect ratios. Since but a slight aerodynamic advan- 
tage is obtained by elliptical or circular tips, especially when the aspect 
ratio of the wing is greater than 5.0, the decided structural advantages 
of the square or modified square tips shown in Fig. 26 make these latter 
types the more satisfactory. A tip that has recently come into favor and 
which has many advantages is one that is absolutely square in plan 
view. The full depth of the wing section is maintained out to the tip, and 
balsa fairing, with a radius at each point along the chord of the wing 
equal to half the wing ordinate at that point, rounds off the wing. Such 
a tip greatly simplifies the construction, since no special short ribs are 
used at the tip and the leading edge does not have to be bent around 
the tip. Such a tip reduces the high speed between one and two miles 
per hour which is negligible in many types of airplanes. The raked tip 
has in the past been largely used because of its supposed efficiency and 
the fact that it afforded aileron area where it was most needed. On the 
other hand, the structural disadvantages of the elliptical or circular tip 
are accentuated in a raked tip. There is also a backward movement of 
the center of pressure which makes these disadvantages even more pro- 
nounced. For these reasons the raked tip is not recommended. 

33. Location of Lift Wires in Inner Bay — It was at one time cus- 
tomary in a staggered biplane to keep the lift wires in the plane of the 
struts. To counteract the large drag loads put into the drag trusses by 
the backward component of the stresses in the lift wires with this ar- 
rangement, external drag wires, carried forward to the nose of the 
fuselage, were used. The location of the wires shown in Fig. 27 is rec- 
ommended as being superior to the old arrangement. The double, front 
lift wire is kept in a vertical plane or brought back slightly. The rear 
lift wires are split. One is carried forward and attached to the fuselage 
either at the point of attachment of the front wire or at the bulkhead to 
the rear of this point. The other rear wire remains in the plane of the 
struts or is secured to a bulkhead so located as to give the wire the de- 
sired backward slope. 

In the high incidence condition when the lift stresses in the front 
truss are large and the anti-drag forces on the wings are a maximum, 
the backward component of the stress in the front wire and in the rear- 
most wire will produce a drag tension in the front spar that will relieve 
the heavy compression there. The forward component of the stress in 
the forward rear wire will partially neutralize this, but since the wings 
are tending to move forward slightly the stress in this wire will be less- 
ened. On the other hand, for the low incidence condition the drag truss 
will deflect backward somewhat, thus relieving the stress in the rear- 
most wire and putting it in the other rear wire which on account of its 
large forward component reduces the compression in the rear truss. 

One reason for not splitting the front wires and bringing one of them 
forward is that to attach a wire to an engine bulkhead would subject 
it to severe vibration. Especially in the case of streamline wires this 

45 



General Considerations 



Art. 33 




Elliptical 



a 



Jquapf 




CiPCULAQ 



CU/2T/55 




Fig. 26. Wing Tips 



^3£ 




Art. 34 Location of Interplane Struts 



would be injurious. Furthermore, locating a front wire at such point 
would result in a large drag compression in the front spar when 
it was most heavily stressed. The practice of using external drag 
wires is not recommended. They make the wing structure indeter- 
minate, and are unnecessary. The drag truss is amply strong enough 
to care for any drag loads, and, in fact, usually does carry the entire anti- 
drag load which in a pursuit airplane is greater than the drag. However, 
if external drag bracing is eliminated, more care must be used in the de- 
sign of the wing anchorage fittings than is necessary when drag wires are 
present. The center section struts, wires, and fittings must also be of 
ample strength. 

34. Location of Interplane Struts — The struts should be so placed 
that the factors of safety of the spars at or near the strut points and in 
the span will be approximately equal. In an airplane of several bays the 
length of the panels decreases from the outer span toward the fuselage 
or center section. The purpose of this arrangement is to reduce the 
bending moment in the spars as the compression increases so as to keep 
the combined stresses nearly equal for the several panels. If there is 
a difference in the factors of safety for the panels, the inner panels should 
have a higher factor as the}' are more vital than the outer panels. It 
should be remembered that too great a strut spacing gives a flat slope 
to the flying wires with a consequent increase of the compression in 
the spars in addition to greater deflections. The combination of in- 
creased compression and deflection produces large secondary stresses. 

As the maximum speed of an airplane increases, the elimination of 
interplane bracing becomes more and more important. In a compara- 
tively low speed airplane an extra bay might well be used. This on a 
high-powered pursuit airplane would add far more to the structural re- 
sistance than would be saved in the weight of the members. In each 
case the problem must be solved by the application of the principle 
of equivalent weight (see Art. 99). The resistance and weight of extra 
interplane struts and wires must be balanced against the increased 
weight of the spars that is necessitated by the greater stresses which 
are the result of a reduction in the interplane bracing. 

From a careful investigation of this subject in which the effect of 
aspect ratio, gap-chord ratio, stagger, and continuity of the spars were 
considered in connection with the monoplane, and the one, two and 
three bay biplane, the following deductions were arrived at and the 
values in Table VI determined. 

I. When the spars are continuous at the center, the bays tend to 
increase in proportion to the total length, and the cantilever to 
decrease. This effect of the continuity of the spars is the most 
important factor in determining the location of the strut points 
in an}' given type of airplane. 

II. A decrease in slope of the lift wires or monoplane struts will 
increase the proportionate length of the cantilevers. It is through 
this change of slope that a change in the aspect ratio or the gap- 
chord ratio affects the economic location of the strut points. In 

47 



General Considerations Art. 34 

any case within the limits of practical designing the effect is not 
great, and where the spars are continuous over the center and 
there is no center span it is negligible. 

III. Where the spars are continuous over the second joint from 
L he left, the economical location of the struts is that which will 
make the factors of safety at this joint and the first joint from 
the left equal. If this second joint is not at the center line of 
the airplane, the proportion of the bay to the right of it and the 
bay to the left of it may be changed considerably without ap- 
preciably changing the strength of this joint. The proportions 
may be chosen so that the bay to the right or the bay to the left 
of this joint will be of the same strength as the joint, or both bays 
may be made stronger. 

IV. Staggering the wings of a biplane does not materially, if 
at all, change the economical location of the strut points. 

V. Owing to the absence of direct compression in the spars of 
the lower wing, it is the upper wing which governs the design. 
A change in the design of the lower wing will cause a change 
in the stress in the outer strut. This will affect the direct com- 
pressive stress in the bay 1 — 2 in the same manner as a change 
in the slope of the lift wires. The effect on the economical loca- 
tion of the strut is negligible for any likely difference in design 
of the two wings. 

VI. If the rear spar be made longer than the front spar, the fac- 
tor of safety at joint 1 of the rear spar will be decreased, and at 
joint 2 and in the bay 1 — 2 will be increased. The probability 
of using this type of construction was considered to some extent 
in recommending proportions. It is suggested that, if the rear 
spar is made the longer, the bays be proportioned on the basis 
of a mean spar with a length equal to that of the front spar plus 
one-third of the difference in the lengths of the two spars. 

VII. The physical properties of the spar section have an in- 
fluence in the economical location of the struts. The effect is 
small with small aspect ratios, but becomes appreciable with 
large aspect ratios such as are found in three bay biplanes. 

In all cases the following data were assumed: 

R.A.F. 15 wing section. 

50 in. chord. 

Front spar 13 per cent of chord back from leading edge. 

Rear spar 67 per cent of chord back from leading edge. 

Net wing loading 7.5 lbs. per sq. ft. 

Location of center of pressure for high incidence condition at 

29 per cent of chord. 
Location of center of pressure for low incidence condition at 

50 per cent of chord. 
Square wing tips. 
Decrease in tip loading as indicated in Fig. 34. 

48 



Art. 34 Location of Interplane Struts 

Spar Properties 

Front Rear 

Area 2.15 2.08 

I 1.94 1.12 

I/y 1.34 0.99 

P 0.95 0.735 

Sections of these spars are shown in Fig. 28. 

TABLE VI— RECOMMENDED PROPORTIONS 

Monoplane: Canti- Bay / Bay 

Without center span — lever 1-2 2-3 

Hinged at center 

Aspect ratio 4.0 34/ 65/ 

Aspect ratio 5.0 35/ 64/ 

Continuous at center 

Aspect ratio 4.0 and 5.0. . . 32/, 67/ 

With center span — 
Continuous throughout 

Aspect ratio 5.0 25 53/ 21/ 

Biplane — Single Bay: 
Without center span — 
Hinged at center 

Aspect ratio 4.0 and 6.0. . . 36/ 63/ 

Continuous at center 

Aspect ratio 4.0 and 5.0... 32/ 67/ 

With center span of 30 in. — 
Hinged at joint 2 

Aspect ratio 4.6 36 64 

Aspect ratio 6.6 37 63 

Continuous at joint 2 

Aspect ratio 5.6 34 66 

With center span greater than 30 in. — 
Continuous throughout 

Aspect ratio 4.0 and 5.3. . . 25 51/ 23/ 

Lower wing only — 
Hinged at joint 2 

Aspect ratio 6.6 32/ 67/ 

Biplane— Two Bay: Entire Entire 

With center span of 30 in. — Bav 3-4 

Hinged at joint 3 

Aspect ratio 7.0 24/ 45/ 30 

Biplane — Three Bay: 

With center span of 30 in. — 
Hinged at joint 4 

Aspect ratio 9.0 20 40 25 15 

Note: — The length of the cantilever in the above table is the 
actual cantilever length of the spars, not the loaded 
length used in sand test. 

49 



General Considerations 



Art. 35 









~T 




1 ,* 








1 






I 




\ 








t 


^ V 






-0 


5-+ 


1 5 








! 




j 












-05- 


./ 










y 




J 




\f 




■» r^ 




<3 




^-^ ^. 






^ 




i 




1 








i 


r 
















/ p/; •*! 








- /.v5" " 








Fig. 28. Sections of Spars Used in Computing Table VI 




Fig. 29 



35. Location of Spars — In the usual type of wing construction two 
spars are employed. They are generally located so as to receive about 
70 per cent of the total load on the wing either at high or at low incidence 
flying conditions. Fig. 29 represents a wing section with two spars whose 
location will be determined under the assumption that each spar re- 
ceives 70 per cent of the load. The center of pressure is assumed to be 
at .29 of the chord from the leading edge at high incidence and .50 of the 
chord from the leading edge at low incidence. 
Low incidence condition: 
.50L — a 
Load on rear spar = XP^ -70 P 



(1) 



.50 L — a = .70b 
a + .70b = .50L 

50 



Art. 36 Torque of Engine 

High incidence condition: 
a + b — .29 L 

Load on front spar = X P = -70 P 

b 

a _|_ b — .29 L = .70 b 
(2) a + .30b = .29L 

Solving (1) and (2) 

a = .14L 
b = .53 L 
An approximate location of the spars is then: 

Front spar — 1/7 of chord from leading edge. 
Rear spar — 2/3 of chord from leading edge. 
These values are suitable for types of airplanes in common use at 
present. For very large airplanes the flying conditions will be such 
that the center of pressure will not go as far back as .50 of the chord. 
The maximum percentage of the load carried by the rear spar will 
therefore become less. Hence, there will be a tendency to move both 
spars forward somewhat in the wing section, the rear spar perhaps more 
than the front. This will help to equalize the loads on the spars. How- 
ever, the rear spar depth will increase, and that of the front spar de- 
crease, as the spars are brought forward. It may be noted that, al- 
though the overhang of the ribs at the trailing edge becomes large, the 
more forward location of the center of pressure at low incidence reduces 
the load on the rear part of the wing. 

It is suggested that reasonable limits for the position of the front 
spar are 10 to 15 per cent of the chord from the leading edge, and for the 
rear spar, 60 to 70 per cent, the lower values being adapted to heavy 
airplanes. 

36. Torque of Engine — With an airplane using two or more engines 
the torque is of some importance. In Fig. 30 let T = torque of engine, 
R = reaction on engine bearers, and d == distance between bearers in 
inches. Let P = horsepower developed by engine at "n" number of 
revolutions per minute. 

33000 P 
T = 



z 7r n 

T 12 X 33000 P P 

R = — = = 63000— (in pounds) 

d 2 7r nd nd 

A simple rule for determining the directions of the reactions is that 
the direction of rotation of the couple. Rd, is the same as that of the 
engine shaft. 

51 



General Considerations 



Art. 37 




Fig. 30. Reactions from Engine Torque 

37. Method of Least Work — The solution of stresses by statics can 
be accomplished only when there are no more than three unknown con- 
ditions. The existence of more than three unknowns makes it necessary 
to employ some method involving the elastic properties of the material. 
Such a method is the well-known principle of least work which states 
that "the deformation of any elastic body under the action of a bal- 
anced system of external forces will be such that the work done in 
causing the deformation, or the resilience of the body, is a minimum." 
This principle is explained in detail in Art. 116. 

38. Properties of Spar Sections — In the case of most spars the top 
and bottom surfaces are seldom at right angles to the vertical axis. The 
cross-section of a typical spar is shown in Fig. 31 by the dotted lines. 
The computations of the properties of such a cross-section may be great- 
ly simplified by assuming the section to be rectangular as indicated by 
the heavy full lines. The depth, d, is the center depth of the true sec- 
tion. The moment of inertia of the spar section will equal 



I 



wd 3 — c (d x 3 -f d 2 3 ) 



12 
When computing the section modulus either the value y or y 1 should 
be used in the expression I/y and not the value of d/2. 

The area may be taken as the difference between that of the main 
rectangle and the areas of the shaded rectangles. A = wd — (d x -\-d 2 )c. 
The computation of the intensity of horizontal shear on any hori- 
zontal plane of the beam requires a property of the cross-section which 
is usually designated by the expression 0/b I. 

Q = moment about the neutral axis of the area on the side of the 
plane (on which the horizontal shear is to be computed) 
away from the neutral axis, 
b = width of the section at the plane under consideration. 
I = moment of inertia of the whole cross-section. 
The application of this formula is given in Art. 20. In computing 
the statical moment 0, the spar may be treated as rectangular, the same 
as in calculating the area or moment of inertia. 

If it is desirable to take into account the fillets in the routing in com- 
puting the spar properties, this may be readily done by roughly com- 

52 



Art. 39 



Eccentricity of Fittings 



puting the area of one fillet assuming it to be a triangle. The moment 
of inertia of the fillets equals their area times the sum of the squares of 
the distances from their centers of gravity to the vertical axis. Even 
without this correction the method given will have an error less than 
1 per cent. 




Fig. 31 

39. Eccentricity of Fittings. — Fig. 32 illustrates a typical wing fit- 
ting on an upper spar. It is evident upon a little consideration that the 
downward pull from the lift wires is carried by bolt A. The true point 
of support for the spar is therefore at point c and not at the center line 
of the strut. The forces acting along the strut and the lift wire have 
eccentricities about point r, and consequently moments are produced in 
the spar. These moments should be considered in the design of the spar 
unless the fitting is so designed that the eccentricities are so small as to be 
negligible. Cases 4 and 5 of the three-moment equations treat this con- 




Fig. 32. Eccentric Wing Fitting 

53 



General Considerations Art. 40 

dition of moments at the panel points (see Art. 172). For the lower 
spar the reaction point is on the center line of the strut. 

40. Maximum Load on Wings — Accelerometer tests have shown 
that the most severe loads on the wings occur in pulling sharply out of 
a steep dive. For a brief period of time the airplane is moving at a 
high, though decreasing velocity, in a path of decreasing radius of curva- 
ture. During the dive the angle of incidence is small, or even negative, 
which means that the center of pressure is far back on the wing. As 
soon as the airplane deviates from a straight path the angle of incidence 
begins to increase and the center of pressure to move forward. The 
maximum dynamic loads, due to the centrifugal force, occur at a high 
angle of incidence, close to the angle giving the maximum lift coefficient 
for the aerofoil. The greatest dynamic load that has yet been recorded 
by an accelerometer in a test is 4.2. It is possible that a load factor of 
5.0 would be obtained in actual combat. The following discussion has 
reference to fast pursuit airplanes, and assumes that 4.2 is the maximum 
dynamic load and that it occurs at a high incidence when the center of 
pressure is about 30 to 31 per cent back from the leading edge. With 
"an ordinary design in which the front spar is located at 12 per cent and 
the rear spar at 67 per cent of the chord, the front truss will carry 2/3 
and the rear, 1/3 of the total load on the wings. Using these propor- 
tions the maximum load that can come on the front truss is 2.8 W and 
on the rear truss 1.4 W , where W is the gross weight of the airplane 
minus the weight of the wings. The necessary factor of safety is then 
arrived at as follows: If the resultant load W is applied so that the front 
truss carries 2/3 W and the rear 1/3 W then the factor to produce loads 
of 2.8 W and 1.4 W is 2.8W/ 2 / 3 W = 4.2. To this is applied a material 
factor or true safety factor of about 2.0, making the total load factor 8.4. 
For the standard static test the required factor for pursuit airplanes 
is 8.5. 

It is evident that the most severe condition for the front truss occurs 
when high incidence is combined with high speed, as in pulling out of a 
dive or in zooming. The corresponding condition for the rear truss will 
be either at high incidence, or in a straight dive when the resultant lift 
on the wings is zero or very small. It cannot occur in level flight at 
high speed, when the center of pressure is far back, because no dynamic 
effect can exist unless the airplane is traveling in a curved path in a ver- 
tical plane. With the high speed condition not more than 1.0 W can 
come on the rear truss, and usually not more than .8 W which would 
correspond with a center of pressure located at about 55 per cent of the 
chord. It is probable that, except for internally braced monoplanes, the 
critical flight condition for the rear truss is the same as for the front 
truss. The case of a straight dive is discussed in Art. 43. In addition to 
these conditions of flight there is that of reversed loading, which is met in 
sharply entering a dive at high speed, or with pursuit airplanes in 
various maneuvers or in actual upside down flight. This condition is of 
importance chiefly in the case of braced or semi-braced monoplanes in 
which it becomes one of the limiting conditions. There are certain other 

54 



Art. 41 High and Low Incidence Conditions 



types of structures for which it would also be a critical condition. The 
load factor for reversed loading should be about .4 that for direct load- 
ing, and in no case less than 2.0. The question of proper load factor 
will be taken up in the appendix to which reference can be made. 

41. High and Low Incidence Conditions — As previously stated, the 
maximum dynamic load occurs at an angle of incidence close to that for 
maximum K y . For the purpose of analysis this angle will be taken as 
that at which the center of pressure reaches its most forward position, 
except for aerofoils with which K y becomes maximum before the center 
of pressure reaches this position. In such cases the angle at maximum 
K y is assumed, and the location of the center of pressure for this angle 
is employed in computations. The values of the angle of incidence and 
of the center of pressure travel are obtained from model tests of the 
particular aerofoil used. Should no tests be available the center of pres- 
sure may be taken at .3 of the chord from the leading edge, and the 
angle of incidence at 14 degs. In determining the direction of the re- 
sultant air load on an aerofoil the L/D at the proper angle of incidence 
must be known. This is obtained from model tests, and should then 
have suitable correction factors, such as the factors for full scale, aspect 
ratio and biplane effect, applied to it. As seen from Fig. 35 this resultant 
air force will have a forward or anti-drag component which may be 
much more severe in certain cases than the backward drag forces in 
other conditions of flight. This is a fact not generally appreciated since, 
as a rule, the drag wires in drag trusses are double, while the anti-drag 
wires are single and since, although external drag wires are commonly 
used, provision is seldom made for external anti-drag wires. Inci- 
dentally, this goes to prove that external drag wires are usually unneces- 
sary, as the drift truss is amply strong enough to carry all drag loads 
either positive or negative. Care should be taken, however, to have the 
incidence wires and -struts in the center section and particularly the wing 
anchorage on each wing capable of transmitting the drag from the 
wings into the fuselage. The resistance of the struts, wires and fittings 
should be calculated and applied as concentrated drag forces at panel 
points. The velocity for which this resistance is found is determined by 
the dynamic or load factor suitable for the type of airplane under con- 
sideration. For pursuit airplanes this may be taken as 4 to 5 and for 
large bombers as \Y\. The velocity, V= V 'FyJV /K y yA . F being the 
dynamic factor, K y the lift coefficient at the specific angle of incidence, 
W the gross weight of the airplane and A the net wing area. For ap- 
proximate computations the resistance of the interplane bracing may be 
neglected, and to compensate for this the value of L/D from model tests 
used uncorrected. It is evident that to underestimate the drag forces 
is conservative in the high incidence condition. The reverse, of course, 
is true for the low incidence condition. This distribution of load between 
wings and along the span is discussed in Arts. 44 to 46. 

In the case of airplanes with a center fuselage and side nacelles, or 
with two fuselages and a center nacelle, the resistance of the nacelle 
should be considered as a concentrated drag load. For neither high nor 

55 



General Considerations Art. 42 



low incidence need the weight of nacelles be considered to have a drag 
component. For both of these conditions the weights of nacelles should 
be taken as concentrated loads applied to the wing cellule and properly 
distributed between the trusses. As just suggested, in an analysis of 
a wing cellule the fuselage or fuselages which carry the tail surfaces 
should always be regarded as reaction points and the forces due to the 
nacelles as applied loads. 

For the low incidence condition the location of the center of pressure 
is determined by the high speed of the airplane at the ground. Know- 
ing this speed the K y can be computed from the formula, K Y = W/AV 2 . 
Then from the curve for the center of pressure travel of the aerofoil the 
location of the center of pressure, for the angle of incidence correspond- 
ing to the K y calculated above, is found. As before, the L/D at the 
same angle of incidence is obtained from the model test and suitably 
corrected for full scale, etc. The resistance of the interplane bracing at 
maximum ground speed is then added to the distributed drag load de- 
termined by resolving the resultant air force on the aerofoil into com- 
ponents normal and parallel to the wing chord. In the low incidence 
condition also, the approximation of neglecting the resistance of the 
interplane bracing and using uncorrected values of L/D can be made, 
though it may not be on the safe side. In case characteristic curves 
of the aerofoil are not at hand the center of pressure may be taken as 
between .45 and .70 of the chord from the leading edge for thin and 
thick aerofoils, respectively, and the L/D as 6. 

Although the low incidence condition is not a limiting one in itself, 
yet, because it can be readily and accurately calculated, it serves as a 
means for proportioning the rear truss. The data upon which the 
computation of the stresses in a straight dive is based are not quite so 
definitely known. But the main reason why the rear truss is not de- 
signed solely for this condition is due to the indeterminate character of 
the cellule under the load that is produced in a dive. In a military air- 
plane, especially one subjected to severe fire, the rear truss must be of 
sufficient strength to carry the entire weight of the airplane when the 
front truss is destroyed, with a safety factor of 2 to 3 depending on the 
size of the airplane. 

The center of pressure for the reversed loading condition is well for- 
ward. Its exact position has not been determined for most aerofoils, but 
it may be taken at .25 of the chord from the leading edge. Information 
as to the angle of incidence and the L/D for this condition is meager. 
It is probably safe to assume that these factors are the same as in the 
high incidence condition, with the exception that the angle of attack 
is negative. Aerofoils at negative angles are much less efficient than 
at positive angles. To counteract this, the angle of incidence is very 
large. 

42. Static Test Conditions for the Wing Cellule — Standard static 
testing as conducted by the U. S. Government at McCook Field has been 
changed so that instead of attempting to combine two conditions of 
flight in a single test, separate tests are made for the conditions of high 

56 



Art. 42 



Static Test Conditions for the Wing Cellule 



and low incidences, and in certain cases, such as braced monoplanes, the 
cellule is tested for reversed loading also. As the front truss is most 
severely stressed in the test for the high incidence condition and the 
rear truss in the test for the low incidence condition, by carefully local- 
izing the failure and repairing the break, the same cellule may be used 
for both tests. No attempt is made to test for the conditions of a 
straight dive. Since the tests are arranged so as to reproduce the condi- 
tions of flight as nearly as possible no separate stress analysis for static 
test need be made. The location of the center of pressure is determined 
as has been described above for the purposes of stress analysis. The 
distribution of load between the wings and along the span will be the 
same as, or equivalent to, the distribution of load for flight conditions. 
Since in static test the load acts vertically, the drag load is obtained by 
inclining the wing chord. The angle of inclination of the wings is such 
that the load will act in the same direction with respect to the wing 
chord as the resultant of the air forces on the cellule. These forces are 
made up of the combined lift and drag forces on the aerofoils and the 
resistance of the struts, wires, and fittings. In general, this angle of 
inclination will be about 7 degs. With the wings in reversed position 
for test, the leading edge will be up for the low incidence and down for 
the high incidence condition. 

In static tests on airplanes with side or center nacelles, the resultant 
of the loads caused by the nacelle, which are made up of its resistance 
and weight, are reproduced by a cable attached to the cellule in a suit- 
able manner. 



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Yo/ues of Max /mum k'ufM.PH un/ts) 

57 



General Considerations Art. 43 

43. Diving — An airplane is assumed to dive at an angle close to the 
angle of zero lift, and to reach its limiting velocity at which the total 
resistance equals the weight of the airplane. Owing to the character of 
the distribution of pressure on a wing at very small angles, shown in 
Fig. 33, there is a large moment which tends to put the airplane into 
an even steeper dive. This moment is counteracted by downward forces 
on the tail sufficient to produce equilibrium. The forces acting on the 
wings are, therefore, downward loads on the front lift truss and upward 
loads on the rear lift truss. The upward loads are greater than the 
downward by the amount of the tail load. In addition to these there 
is the drag of the wings, struts, and wires. 

From wind tunnel tests on aerofoils, moment coefficients for the 
wing section used can be obtained. The total diving moment then 
equals M = K m pcSV 2 , D=W = K d pSV 2 . From these relations 
M = K m c W/K A . 

W = gross weight of airplane in lbs. 
P = density of air = .07608 lbs. per cu. ft. 
S = total area of wings in sq. ft. 
c = chord length in ft. 
V — velocity in ft. per sec. 
K m = moment coefficient in absolute units. 

K d == coefficient of resistance of entire airplane in absolute units. 

Values of K d based on full flight performance data are listed below 

for various airplanes. From these, the value of K d for any airplane may 

be estimated by comparing the airplane in question with the types in 

the table. Airplane K d in absolute units 

Martin Bomber 0375 

JN-4 D2 0323 

DH-4 0297 

U.S.D.-9A Bomber 0297 

Fokker D-7 0272 

VE-7 0249 




Fig. 33. Pressure Distribution on Wing in a Dive 

58 



Aft. .43 Diving 

Ordnance D 0242 

SE-5 0236 

MB-3 0212 

U.S. XB-1A 0206 

V.C.P.-l 0200 

Curves of values of K m are given for K y = .0000 and .0003 (M.P.H. 
units). These curves are averages of the K m , at these values of K y of a 
large number of different aerofoils with maximum K y s ranging from 
.0022 to .0039. Although in individual cases the K m of an aerofoil may 
not fall exactly on the curves, the error will be small, especially in the 
lower portion of the curves which is the most important part since on 
airplanes that will dive very steeply, speed sections, such as the R.A.F. 
15, with low maximum K y 's will be used. Furthermore, in case an aero- 
foil having a high maximum K Y is employed, the K d of the airplane will 
be large. 

When the diving moment, given by the formula M = K m cW/K A , 
has been computed, the load P on the tail plane may be found. If it is 
assumed that the centers of resistance and of thrust are at the center of 
gravity of the airplane, which is reasonable and conservative, P = M/L, 
L being the distance from the center of gravity of the airplane to the 
resultant of the load on the tail plane. The load on the trusses mav now 
be calculated as follows: 

F l = load on front truss. 

F 2 = load on rear truss. 

h 1 = distance from front truss to the resultant of the load on 

the tail plane. 
L 2 = distance between front and rear trusses. 
Taking moments about the front truss, 
L t P = L, F, 
L,P 

F 2 = = F t + P 

L 2 
With F ± and F 2 known, the loads per inch run on the trusses can be 
computed and the stresses determined in the usual manner. It should 
be remembered that the loads on the front truss act downward and 
those on the rear truss upward. In the usual type of wing construction 
in which incidence wires are used, the latter will be severely stressed, and 
will tend to neutralize the truss stresses, thus considerably increasing the 
factor of safety. The bending moments on the spars produced by the 
distributed load and the stresses due to them will not, however, be 
affected by the incidence wires. A factor of safety of 1^4, neglecting 
the incidence wires is ample for this condition of diving. 

In the internally braced monoplane there are, of course, no inci- 
dence wires tending to equalize the truss loads, and if such an airplane 
can be made to dive at an angle of zero lift much more severe stresses 
will be induced than in the case of an ordinary biplane, unless the wing- 
is deep enough to permit effective internal bracing between the spars. 
In addition to the stresses caused by the up and down loads on the 

59 



General Considerations Art. 44 



lift trusses, there are large drag stresses which occur in a limiting dive. 
The proportion of the total drag that is made up of the drag of the 
wings and the resistance of the interplane struts, wires, and fittings 
varies from 45 to 60 per cent of the gross weight of the airplane. In an 
extremely clean wing structure with a speed aerofoil, a minimum of 
interplane bracing, and with streamline wires, the lower percentage is 
applicable. It is seldom that the wing drag exceeds 60 per cent, 
unless the wing loading is very light. For the U.S.D-4 it amounts to 
55 per cent. To obtain the net drag on the wings, their weight, which 
may usually be assumed as 15 per cent of the total weight of the air- 
plane, must be subtracted, leaving 30 to 45 per cent of the gross weight. 
If side or center nacelles are used, their drag must be added to this. 
Data relative to coefficients of resistance of nacelles, especially engine 
nacelles, are unsatisfactory. It may be said that in a limiting dive the 
resistance of an engine nacelle averages about 30 per cent of its weight, 
and if the nacelle carries a gas tank, also, 20 per cent of its weight. The 
weight of the nacelle acts vertically downward, and that component 
parallel to the line of flight must be subtracted from the nacelle re- 
sistance to give the net drag, or rather anti-drag. 

44. Distribution of Load Between Wings — Reliable information on 
this subject is meager. The quantitative effect of various factors such 
as stagger, gap-chord ratio and angle of incidence has not been deter- 
mined. Several tendencies may be pointed out, however. 

At both high and low incidence conditions positive stagger decreases 
and negative stagger increases the relative lift on the lower wing of a 
biplane. At very high angles of incidence the lower wing becomes more 
effective than the upper wing. This is because the upper wing reaches 
its burble point sooner than the lower and so loses its efficiency. 

In a biplane with a gap-chord ratio between 0.8 and 1.2, and no 
stagger, the load per sq. ft. on the lower wing may be taken as 90 per 
cent of the load per sq. ft. on the upper wing for the high incidence con- 
dition, and as 95 per cent for the low incidence condition. 

When there is 50 per cent positive stagger, these values should be 
decreased to 80 per cent and 75 per cent for high and low incidence, 
respectively. For 50 per cent negative stagger they increase to 110 per 
cent and 125 per cent. 

With a triplane the loads per sq. ft. on the middle and lower wings 
may be taken as 75 per cent and 85 per cent, respectively, of the load per 
sq. ft. on the upper wing in the high incidence condition. For the low 
incidence condition both of these values become 80 per cent of the load 
per sq. ft. on the upper wing. 

In a standard static test the load would be distributed in accordance 
with the proportions given above. 

45. Distribution of Load Along Span and Chord — The following 
assumptions regarding the location of the center of pressure will be 
made as representing a close average for most wing sections of the gen- 
eral character of the R.A.F. 15. The location is given in per cent of the 
chord from the leading edge. In the case of a few wing sections the 

60 



Art. 46 



Distribution of Pressure Along Wing Tip 



center of pressure movement is much greater than is indicated below: 

Low incidence =50 per cent of chord 

High incidence =30 per cent of chord 

Static test (direct loading) . . . = 30 and 50 per cent of chord 
Static test (reversed loading) . = 25 per cent of chord 
For unusual wing sections in which the center of pressure travel is 
much greater or less than is indicated by the first two of these values, 
proper percentages obtained from wind tunnel tests should be used. 

In static tests the load along the span is constant. The effect of 
rounded tips and of reduced loading on tips is allowed for by decreas- 
ing the loaded length on the cantilever span so that the cantilever 
moment will be the same as the moment of the loading treated in i\rt 46. 
46. Distribution of Pressure Along Wing Tip — No extensive series 
of experiments have been carried out with a view to determining the 
variation in pressure on the wing tip. Experiments conducted at the 
Royal Aircraft Factory resulted in the curve given in Fig. 34, which is 
a fair representation of the diminution of pressure at the tip. 

In order to simplify computations the distribution is taken as shown 
by the dotted line. At a distance from the wing tip equal to the 
chord the load per inch of run is reduced to 0.9 of the maximum and at 
a point 1/2 of the chord length from the tip the load is reduced to 0.7 
of the maximum. Whenever the cantilever length is less than the chord, 
the maximum load is carried out to the strut point to facilitate the cal- 
culations of the moments and shears on the spars. Since the load to be 
carried by a wing is a definite amount, it is evident that the maximum 
load per inch of run is greater when allowance is made for the decrease 
toward the tip than when the load is uniform over the entire span. 

















^^ 


f^ 


nfT 


"" 





















i 
















^^ 


^'"\ 


















(J .6 
< 


., - 


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9i 




















































































































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.c 


J 


4 




,t 




t 


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/ 





/. 


j 


M 



A/umber of T/mcs Chord from Wt'ng T/p. 
Fig. 34. Distribution of Load on Wing Tip 

47. Distribution of Load Between Spars — For the condition of load- 
ing under consideration, the location of the center of pressure, and hence 

61 



General Considerations 



Art. 47 



of the resultant force on the wing, is first determined. The proportion 
of the total load carried by the wing that goes into each spar is inversely 
proportional to the respective distances of the spars from the resultant 
force. From the gross load supported by the wing must be subtracted 
the weight of the wing cellule, in order to get the net load on the wing. 
For a small airplane with a low safety factor the weight of the wing- 
cellule may be as low as .9 lbs. per sq. ft. of wing surface, while with 
large airplanes of the bombing type the weight will seldom be less than 
1.5 lbs. per sq. ft. For a usual design of a medium-size airplane this 
weight may be taken as about 1.1 lbs. per sq. ft. 

The loading on the wing tip is reduced as described in Art. 46. The 
running loads on the front and rear spars are then calculated by the 
following: formulas: 



W 



+ 



.9W,!, 



+ 1.0w f l 



.7w r l r + ^Wrl, + 1.0w r l + .7W f l, 

w f = w r -p f /p r 

W = total net load on wing. 

w f = normal running load on front spar. 

w r = normal running load on rear spar. 

p f = per cent of W on front spar. 

p r = per cent of W on rear spar. 

1 = length of front and rear spars with normal load. 

\ x = length of front and rear spars with .9 normal load. 

l f = length of front spar with .7 normal load. 

l r = length of rear spar with .7 normal load. 

li+lt £ = cantilever length of front spar. 

1-l— j— l r = cantilever length of rear spar. 
As a rule, no reduction of load on the rear spar due to cutouts at 
the fuselage is allowed for. This method will give a correct distribution 
of load with square or only slightly rounded wing tips. If the wing tip 
is given a large rake, the load on the rear spar given by this formula 
is low because with a raked tip the center of pressure moves back. Yet, 
except in extreme conditions, this method will give satisfactory results. 
No aileron loads, in addition to the normal load on the rear spar, are 
considered. Ailerons would be used but little under the flight conditions 
which produce maximum stresses in the rear spars. An exception is 
made in the case of a balanced aileron, part of which projects beyond 

■Tongenf/o/ Component 



N / 


/e'r ^~~ r ^~~^~-^_ 


v/ hi 

5/ '* 

M 

Dr/ft*/ .^V 

i/AS 

T 


Hor/zonto/S 

Fig. 35. 
Resolution of Forces 



62 



Art. 48 Resolution of Lift and Drag Forces 



the contour of the wing proper. On this portion of such an aileron a 
loading equal to .7 of the normal wing loading is assumed. All of this 
load is carried by the rear spar. 

48. Resolution of Lift and Drag Forces — The resultant pressure on 
a wing is composed of a vertical force, called "lift," and a horizontal 
force, called "drag." The ratio of lift to drag (L/D) determines the 
inclination of the resultant force. Fig. 35 is a resolution for the high in- 
cidence condition in which the angle of incidence is 12 degs. positive and 
L/D equals 8. The resultant is resolved into a component normal 
to the wing chord and a component parallel to the wing chord. The 
normal component produces stresses in the lift trusses while the tan- 
gential component produces stresses in the drag trusses. The drag is 
always laid off parallel to the direction of the wind and the lift per- 
pendicular to this direction. 

The vertical component of the resultant equals the weight of the 
airplane, but for designing purposes it is sufficiently accurate to take 
the resultant itself as equal to the weight. In the sand test loading 
the resultant is vertical, and, when the angle of inclination of the wing 
chord is 7 degs., the component normal to the wing chord is .992 of the 
resultant and the component parallel to the chord .122 of the resultant. 

49. Moments, Shears, and Reactions — When the running loads on 
the spars have been determined, the next step is the calculation of the 
moments on them at their supports. If a spar is continuous over sev- 
eral supports or if the loading is not uniform, the necessary three 
moment equations can be found in Art. 172. Care should be taken 
not to assume a spar continuous unless it is truly so. All properly con- 
structed splices can be considered continuous as well as joints made 
with fairly long metal sleeves. There are few types of wing anchorage 
fittings, however, that justify the assumption of continuity, even though 
they may theoretically do so before the close fit of the parts is destroyed 
by the wear and tear of service. From the moments at the supports 
the shears and reactions of the spars ma}' be computed by the methods 
of Art. 19. If in the truss under consideration there are no moments 
from eccentric strut loads or wire pulls, then the next step is the cal- 
culation of the bending moment in each bay at the point of zero shear. 

50. Truss Stresses — Both lift and drag trusses are assumed to be 
pin-jointed structures for the calculation of the truss stresses. No ac- 
count is taken of the effect of deformation of the truss under load. All 
truss loads are considered as concentrated loads applied at the panel 
points. With a lift truss these loads are the reactions from the spars, 
calculated as described above. When any weights, such as engine 
nacelles, are located on the wing their effect in relieving the lift stresses 
must be allowed for. The presence of incidence wires connecting the 
front and rear trusses renders them indeterminate. Therefore, except in 
considering cases in which lift wires are shot away or in which the load 
on one truss is up and on the other down, as in a steep nose dive, the 
effect of incidence wires is neglected, unless, as in a static test ten- 
siometer measurements have been made of the stresses in them. Either 

63 



General Considerations Art. 51 



graphical or analytical methods can be used in calculating truss stresses. 
If care is taken it is believed that the graphical method is somewhat 
more satisfactory. 

51. ' Eccentric Strut Loads and Wire Pulls — After the lift truss 
stresses have been determined the effect of eccentric struts loads and 
wire pulls can be computed by means of the three moment equations in 
Art. 172. This effect is most serious in single bay airplanes with the 
spars hinged at the fuselage or center section. If the spars are con- 
tinuous over two or more points of support the stresses from eccen- 
tric loads will not largely alter those produced by the direct uniform 
load, especially within the spans. With eccentric wire pulls, the mo- 
ments at the strut points due to the wire eccentricities change from 
positive to negative. However, in calculating the fiber stresses at these 
points, the stress due to the maximum moment should be combined 
with that due to the maximum compression, since the maximum values 
of each type of stress occur simultaneously. In a single bay airplane or 
in the case of unusually large eccentricities the reactions produced by 
the eccentricities should be computed, and the reactions from the direct 
air loads modified accordingly before obtaining the final values for the 
lift truss stresses. After the moments and shears from the direct air 
loading and the eccentric loads are combined, the points of zero shear 
should be determined, and then the bending moments at these points 
calculated. 

52. Deflection Moments — When the stresses in the lift and drag 
trusses and the moments and shears from the direct air load have 
been calculated, and after any modification in them necessitated by 
strut or wire eccentricities has been made, the deflections of the 
spars, at the final points of zero shear within the bays in which the 
spars are under compression, should be computed from the formulas 
for deflection in Art. 172. It is unnecessary to obtain separately the 
deflection due to eccentric wire pulls, for if the moments and shears at 
the strut points due to eccentric wire pulls are combined with the 
moments and shears from the air load, and these combined values are 
substituted in the formula for deflection due to uniform load, the de- 
flection obtained will be the resultant of the deflection produced by the 
eccentric wire pulls and the uniform load. If, however, there are pres- 
ent eccentric strut loads also, the deflections caused by them should 
be computed separately. In order to obtain the deflection it is necessary 
to know the moment of inertia of the spar, and hence if it has not been 
designed some approximate size must be assumed for the preliminary 
calculations. When the final moment of inertia is known the deflection 
should be corrected. The product of this deflection and the compres- 
sion in the span gives the deflection moment which is added to the bend- 
ing moment in the span, previously obtained. Because the panels of 
the drag truss are only half as long as the panels in the lift truss there 
will be a change in the compression near the center of the span. The 
maximum compression is the one to use in finding the deflection mo- 
ment; i. e., if the stress in the drag truss is tension, then the smaller 

64 



Art. 53 Stress Table 



value of the tension should be subtracted from the lift compression, 
while if the drag stress is compression the larger value should be added 
to the lift stress. 

53. Stress Table — The tabulation of spar properties, moments, com- 
pressions, unit stresses, and factors of safety given in Table XI is in the 
most satisfactory form for use. The values of Pd in column 8 represent 
the product of the compression and the deflection each calculated for 
unit load factor. Since this deflection moment is the product of these 
two factors, each of which vary as the load factor, it varies as the square 
of the load factor. Hence, the factors of safety of spars in the spans 
under compression must be determined by the method of trial. Be- 
fore the deflection moment as given in column 8 can be added to the 
main moment in column 7 it must be multiplied by the factor of safety 
for the spar in the bay under consideration. By doing this the fact 
that the deflection moment varies as the square of the load factor is 
taken into account. 

54. Unit Fiber Stresses — In airplane design the stresses employed 
are based on ultimate strengths rather than elastic limit fiber stresses. 
There are two main reasons for this. First, airplane structures are very 
frequently subjected to static tests which are carried to destruction. 
They are designed to have a definite load factor under these conditions. 
Second, airplane structures are unique in that the extreme loads im- 
posed upon them are of very short duration. In wood, especially, little 
injury is done by exceeding the elastic limit or even approaching the 
ultimate strength, provided the load is removed within a few seconds. 

Tables XXXIX and XL of the appendix gives the physical properties 
of steels and various non-ferrous metals. These tables were compiled 
from standard government specifications. Ultimate stresses for steel 
columns and struts can be obtained from the parabolic formula given 
in Art. 24. This formula is regarded as more satisfactory than the 
Rankine or the various straight-line formulas used in structural work. 
This is particularly true for high strength steels and for materials on 
which few data regarding column strength are available. Attention 
should be called to the fact that with ductile materials the ultimate com- 
pressive strength for struts will not be greater than the yield point of 
the material. 

In thin walled tubes the ultimate strength may be further limited 
by so-called "crinkling action," by which is meant local crumpling of 
the tube wall. This can be avoided by using tubes in which the ratio of 
wall thickness to tube diameter is greater than about .06. When the 
ratio is less than this the crinkling stress falls below the yield point 
of the material and the design may be limited by this stress rather 
than by the ultimate strength computed from the parabolic formula. 
The curves in Fig. 36 are suggested by Mr. Barling as giving a reason- 
able reduction in ultimate strength due to crinkling action. In con- 
nection with thin walled tubes it is recommended that when the tubes 
are to be welded, the gage should not be less than No. 20 (.035 in.) 
although skilled welders can weld No. 22 gage material satisfactorily. 

65 



General Considerations 



Art. 54 



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1 . /•c/j _. ^ i i 4. K ^ 








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7 








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i *" ^ 






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.. __+ ^ ... .+_ .... _.. ^ .. ._ ._ .... _. _. _. ... _. 






: i"±± ' . ~ ' X -5$ ~ i" S ""-" " - - x"js _ 












\ o \ a 








' - j— -"-- - " - ■+" rv; p T+- i -i- ft •<■ \ -■--■■ - + -+ vs 


X 1 - I ; | \l V V^ 






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aet-ivr mo 



A similar reduction in ultimate torsional stress because of thin walled 
tubes is treated in Art. 149 and a curve showing this reduction for me- 
dium carbon steel is given in Fig. 153. For alloy steels the curve would 
probably be similar, with all the values proportionally increased. 

In the appendix, Art. 174, a rather complete discussion is given of 
stresses in airplane woods including the usual method for calculating the 
ultimate allowable stress in spars carrying both axial and lateral loads. 

66 



Art. 54 Unit Fiber Stresses 

For high ratios of L/p, i. e., greater than about 90, this formula will not 
give safe results as it does not make sufficient allowance for the column 
action which is present when L/p is large. Experimental work on com- 
bined bending and compression is now in progress. Until it is com- 
pleted the following discussion and methods are suggested and it is be- 
lieved will give satisfactory results. 

In a spar subjected to lateral and axial loads the deflection may be 
regarded as being made up of two parts, a primary deflection produced 
by the lateral load and a series of secondary deflections due to what may 
be termed column action. The first type of deflection can be computed 
for any condition of loading and for continuous as well as simple beams 
by means of the formulas and methods given in Arts. 172 and 173. The 
rest of this discussion, however, applies to simple beams. The com- 
pression in the spar at any point has an eccentricity equal to the de- 
flection of the spar at that point. Thus the spar has imposed upon it 
a bending moment which causes the first secondary deflection, as it may 
be called. In a similar manner bending moments are produced by the 
compression acting through this secondary deflection, and another de- 
flection occurs. This last deflection in turn causes further deflection, 
but each deflection is smaller than the preceding one and the ratio be- 
tween successive deflections remains constant. These deflections, there- 
fore, form a series in geometric progression and ma}' be readily summed 
up, once the geometric ratio has been determined. An example will be 
worked out illustrating the method. 

Assume a strut loaded as shown in Fig. 37. The primary deflection 

of this strut under the lateral load of 2.0 lbs. per in. = 

5 wL 3 5 X 200 X 100* 
o / 1 

384 EI 384X1,600,000X2.67 
The maximum bending moment caused by this deflection = m 1 = 
2000 X -61 = 1220 in. lbs. Since the error in assuming the elastic curve 
of the strut to be parabolic is negligible, the bending moment on the 
strut due to the deflection has a parabolic variation and is the same as 
the bending moment produced by a uniform load. The value of 
this hypothetical load is determined bv the equation 1220 in. lbs. = 
zvL 2 wXlOO 2 

= or w = .975 lbs. per in. The first secondarv deflection 

8 8 

due to this uniform load can be calculated from the primarv deflection 
by direct proportion: 

w 2 .975 

8 2 = 8 x x — = .61 X = .297 in. 

w, 2.00 

In a similar manner this secondary deflection produces bending moment 
along the beam which results in a second secondarv deflection. 

w,Xl00 J .475 

m = 2000X-297=594 in. lbs. = ; w 3 = .475 ; 8..=.297x 

8 .975 

67 



General Considerations 



Art. 54 



= .145 in. The ratios of 8 2 to 8 L and 8 3 to 8 2 are equal: .297/.61 = 
.145/.297 = .487, which is the geometric constant. The moment due to 
the entire series of deflection = ra n = 1220 in. lbs. (1.00+.487+.487 2 + 

1 X m 1 1220 

.487 3 + 487 n ~ i ) = = = 2380 in. lbs. for a load 

(1.00— r) (1.00— .487) 

factor of unity. As explained in Art. 53 this deflection moment varies 
as the square of the load factor and must therefore be multiplied by 
the load factor before it can be added to the direct moment caused by 

2.0X100 2 

the uniform load. This latter moment equals = 2500 in. lbs. 

8 
Assume a load factor of 2.75. 

M r = (2500 + 2.75 X 2380) = 9040 in. lbs. 
M t • y 9040 X 1 



ft 





I 




2.67 






p 

A 


= 


2000 

8 


= 250 lbs. 


per sq. 


in 








= 3 640 lbs 


. per sq. 


in. 



If 10,300 lbs. per sq. in. equals the modulus of rupture of spruce and 
5,500 lbs. per sq. in. its ultimate compressive strength, the safety factor 
for this beam is calculated as follows: 



3390 



f„ 



F.S. 



(10,300-5,500) + 5,500 = 9,960 



3640 
9960 



= 2.74 



3640 



zo oo* f/?/////////////////////z/?//mw///////^/i{ZR 



/OO 1 



/oo 



^ooo^ 



/oo 



Fig. 37. Laterally and Axially Loaded Struts 



It should be noted that in computing the ultimate allowable stress 
when the total deflection is known the ultimate compressive strength 
is not reduced to allow for column action but is its full value. In order 
to determine the geometric constant it is unnecessary to compute more 
than the primary and first secondary deflections. Cases will arise in 
which the value of this geometric constant will be greater than unity. 
The series then diverges instead of converging, indicating an indefinite 

68 



Art. 54 Unit Fiber Stresses 



deflection. This means that the beam is in an unstable condition, and 
either its section must be increased or the imposed load decreased. 

One valid objection to this general method is that it will not give 
entirely correct ultimate loads because the elastic properties of the ma- 
terial do not remain constant when the elastic limit has once been 
passed. The deflections tend to increase faster than the formula would 
show. 

With a beam continuous over one or more supports the correct 
primary deflection is calculated by formula, but it is impracticable to 
compute exactly the secondary deflections resulting from column 
action. If, however, in determining these secondary deflections the spar 
is assumed to be a simple beam the same procedure that is given above 
may be used. This assumption will make the deflections somewhat 
greater than they should be, which will offset the tendency of the de- 
flections to be larger than calculated after the elastic limit has been 
passed. 



69 



CHAPTER III 
Wing Stress Analysis 

55. Preliminary Data — The VE-7 Training airplane has been se- 
lected for analysis, and the trusses, wing plans, and spar sections are 
shown in Figs. 38 to 41 inclusive. The data which can be taken di- 
rectly from the plans are: 

Chord 55^ in. 

Gap 56 in. 

Gap-chord ratio 1 .01 

Span 410.6 in. 

410.6 

Aspect ratio = 7.4 

55.5 
The areas of the wings were determined by carefully laying out the 
wings to a scale of 1 in. to 1 ft. and dividing the total area up into 
smaller sections whose area could be readily computed. 

Area of upper wing. .. 151.0 sq. ft. 
Area of loAver wing... 138.8 sq. ft. 

Total area of wings... 289.8 sq. ft. 

The weight of the airplane fully loaded is 1937 lbs. 

The analysis will be carried through for the high incidence condition 
only, as this case will bring out the methods of stress analysis. The 
other conditions are repetitions of this one with differences of loadings. 

56. Location of Center of Pressure — The center of pressure for high 
incidence will be assumed to be at 29 per cent of the chord from the lead- 
ing edge, or .29x55.5=16.10 in. In the same manner, for high speed 
the center of pressure is 50 per cent or 27.75 in. from the leading edge. 

57. Distribution of Loads Between Spars — The front spar is 8.75 
in. from the leading edge. For high incidence the center of pressure is 
7.35 in. (16.10 — 8.75 = 7.35) from the front spar. Since the spars are 
27.375 in. apart, the proportion of load carried by the rear spar 
equals 7.3 5/27.375=. 269. The front spar will receive the remainder, 
or 1.000 — .269 = .731 of the total load. The distribution for low inci- 
dence is computed in the same way. 

DISTRIBUTION BETWEEN SPARS 

Loading % Load on % Load on 

Front Spar Rear Spar 

High incidence 73 27 

Low incidence 30 70 

58. Distribution of Load Between Wings — The proportion of load 
carried by the upper and lower wings has been changed since the present 
analysis was completed. Therefore, the old proportions will be used 
instead of the later values recommended in Art. 44. 

A =151.0 sq. ft. B = 138.8 sq. ft. 

70 



Art. 58 



Distribution of Load Between Wings 






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Distribution of Load Between Wings 







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73 



Wing Stress Analysis 



Art. 58 









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Fig. 41. Spar Sections 
74 



Art. 59 Load on Wings 

The weight of the wings will be taken as 0.9 lbs. per sq. ft. in deter- 
mining the value of the net weight of the airplane, W. Then W equals 
1937 _ 289.8X-9=1937 — 261=1676 lbs. 

Substituting these values in the formula — 

11 

— X 151.0 w+ 138.8 w= 1676 lbs. 
9 

w= 5.19 lbs. per sq. ft. on lower wing. 

11 

— w== 6.34 lbs. per sq. ft. on upper wing. 
9 

59. Load on Wings — The load on the upper wing is 

151.0 X 6.34 = 956 lbs. (net) 

151.0 X (6.34 + .90) = 1093 lbs. (gross) 

The load on the lower wing is 

138.8 X 5.19 = 720 lbs. (net) 

138.8 X (5.19 + .90) = 845 lbs. (gross) 

60. Distribution of Pressure on Wing Tips — Fig. 42 is a plan view 
of the wing and gives the distribution of load per inch of run. As ex- 
plained in Art. 46, the maximum load is carried out to the strut points 
for ease of computation. 

61. Load per Inch of Run — The loads for each spar are given in 
Fig. 43. The method of calculating these loads will be illustrated by 
taking the upper wing spars as examples. Let w = maximum load per 
inch of run on rear spar. By reference to Fig. 43 it will be seen that at 
the strut point the load drops to S)w and extends for 14 in. toward the 
wing tip. Then the load drops to !zv and continues to the end of the 
spar for a distance of 26 in. At high incidence the rear spar carries 27 per 
cent of the total load, and the front spar 73 per cent. Hence the front 
spar carries 73/27 = 2.70 times the load on the rear spar. The max- 
imum load per inch of run for the front spar is therefore 2.1w. As with 
the rear spar, the load drops to .9 of the maximum at the strut point, 
or to .9 X 2.70^ = 2.43 w, and at 14 in. out from the strut point it drops 
to .7 X 2.70w = 1.89cc. Since only one-half of the wing is shown in 
Fig. 43 the sum of these uniform loads must equal one-half the load 
on the wing. 

Rear Spar 164 X 1.0 w = 164.0 w 
14 X -9 w — 12.6 w 
26 X -7 w= 18.2 w 

Front Spar 164 X 2.7 w = 442.8 w 
14x2.43w= 34.0 w 
18 X 1.89 w= 34.0 w 



Total 705.6 w 

75 



Wing Stress Analysis 



Art. 61 



From Art. 59 the net load on upper wing is 956 lbs. 

956 



Then 705.6 w 



= 478 lbs. 



w = .68 lbs. per in. of run 

Substituting this value in the above expressions for loads per inch 
of run the true values may be obtained. The loads for the lower wing 
were obtained in the same manner. 



/<? 



~<L Upper Wing 

— Wing Butt flower) 



32 



4/£* 



^vv^^w^v - vo-y- ■ 



/d Fron/ 



£6 &eor <5por 



?'? 



27? 



/64 Upper 



jWWW k' ■ .' W.I W W*W-V- *■' 



/<?<?" Lower 



/.O 



/4 



27 i 



SS chord 



/6 4 Upper 



/4d L ower 



Z dropped for con sentence 

Fig. 42. Distribution of Load on Wing Tip 



76 



Art. 61 



Load per Inch of Run 



.7* \ sn 

*7*/' 



£6 



XL 



P 



e.<?3# 



70.3 " W) 36J 7 p" 

Re* 12 Upper 5p*/2 



Z.7QW = /.63V 



S'l6 



27% 



" J© 




Front Up pep 5p/?p 



t\ 



H/--.S6 /' 



.7* W/- 


f>m 


'39*7'\ 


p 76S 1© 70S' 
REAR LOWE/2 5 PAR 


P 



27% 



£<J3h/ 



e.70 »/ - /.so*/ " 



,69k\'-36*/ 



/OS*, 



/6 



14- 



Pm 



J0 755™ ~~J@ 7 as" j@ 
FeoNT Low e.g Jp/?e 

Fig. 43. Running Loads on Spars 



13% 



77 



Wing Stress Analysis 



Art. 62 



62. Moments and Shears in Front Upper Spar — The connection of 
the wing spars to the center section spars is shown in Fig. 44. This joint 
is capable of transmitting bending moment, and therefore the spars in the 
upper wing will be assumed as continuous across the center section. In 
the lower wing, however, the spars are pinned at the fuselage. The strut l* ( 
points are assigned station numbers in Fig. 45. The outer strut point is I 
station 1, intermediate strut point, station 2, and so forth. The com- 
plete calculations for moments and shears are given below. The handling 
of the three-moment equation is explained in Art. 22. The important 
relation between the shears and moments in a span (discussed in Art. 
19) is used frequently in this work and should be well understood. 



Hi 

m 




Fig. 44. Joint in Upper Spar 



Front upper spar: 
m 1 = (18 X 1-28) X 23 + (14 X 1-65) X 7 = +692 in. lbs. 

Apply the three-moment equation, Case 2, to spans 1 — 2 and 2 — 3 

1 83 

(78.5x690)+2m. > (78.5+58.5)+58.5m 3 = 

4 
(1) 274 m 2 + 58.5 m 3 = 258740 

Apply the three-moment equation, Case 2, to spans 2 — 3 and 3 — 4a 
noting that m 4 = m s because of symmetry about the center line of the^ 



(78.5) 3 +(58.5) : 



airplane: 

58.5 m 2 +2 m 3 (58.5+54.5) + 54.5 m 3 = 



1.83 



(58.5) 3 +(54.5) 



(2) 58.5 m 2 +280.5 m 3 = 165650 

Eliminate ra 2 between (1) and (2). Divide (1) by 274 and (2) bv 
58.5. 

(3) m 2 +.2135m 3 = 944.3 

(4) m 2 +4.7949m 3 = 2831.1 
Subtracting (3) from (4) 

4.5814 m 3 = 1886.8 

m 3 = +412.0 

ml + .2135X412.0 = 944.3 

m ; = 944.3 — 88.0 =+856.3 

s. x " = — 18X1.28— 14X1.65=— 46.1 

1.83X(78.5) 2 
78.5 s,+692H = m 2 = +856.3 



78 



Art. 63 Moments. Shears and Reactions 

5474 

78.5 

S _ 2 = _(1.83X78.5)— (— 69.7)=— 1437+69.7 = — 74.0 

1.83 X (58. 5; 2 

58.5s 2 +856.3 =-412.0 

2 

3575 
s 2 = = —61.1 

58.5 

s-3 = — (1.83X58.5) — (—61.1) =—107.1-61.1 =—46.0 
L8.3X54.5 

r 1 = s 1 + s. J =—69.7 — 46.1 =—115 
r 2 = s 2 -f s_ 2 = —61.1 — 74.0 = —135.1 
r a = g 3 + 8 „ 3 = _49.9 — 46.0 = — 95.9 



Total = 346.8 lbs. 

Total load on half spar = 23.0+23.1+300.6=346.7 lbs. (check) 
The maximum moments in the spans occur at the points of zero shear. 
Consider span 1 — 2 

69.7 

Distance from station 1 to point of zero shear = = 38.1 in. 

1.83 
1.83X(38.1) 2 

m,.,= -692 (69.7/38.1) =—638 in. lbs. 

2 
Considering span 2 — 3. the distance from station 2 to point of zero 
61.1 

shear equals = 33.4 in. 

1.83 

L83X(33.4) 2 

m..... = -856— (61.1X33.4H = —164 in. lbs. 

2 
Considering the span 2 — 3. the point of zero shear occurs at the 
center of the span. 

L83X(27.25 - 
m 3 _ 3 = +412 — 49.9X27.25 = —268 in. lbs. 

Fig. 45 gives the curves of the moments and shears. It will be noted 
that the points of maximum moment occur where the shear equals zero 
or changes from plus to minus. 

63. Moments, Shears and Reactions — The computations for all 
spars follow the example given in Art. 62. Table 7 is a summary of the 

79 



Wing Stress Analysis 



Art. 63 




Art. 64 Front Lift Truss 

results for the four spars. They do not include the effect of eccentric 
loading, which will be treated in a later article. 

TABLE VII 
MOMENTS, SHEARS AND REACTIONS 
Front Lower Rear Lower Front Upper Rear Upper 



s_. 



m s 

m, 
m, 
m 2 

m. 



92.4 lbs. 39.3 lbs. 115.8 lbs. 48.8 lbs. 
128.8 45.9 135.1 47.2 
40.1 15.4 95.9 36.4 

— 37.8 — 17.1 — 46.1 — 20.7 

— 54.6 — 22.2 — 69.7 — 28.1 

— 63.1 — 21.8 — 74.0 — 25.3 

— 65.7 — 24.1 — 61.1 — 21.9 
_ 40.1 _ 15.4 — 46.0 — 17.9 

0.0 0.0 — 49.9 — 18.5 

+ 567 in. lbs. +323 in. lbs. +692 in. lbs. +390 in. lbs 

+900 +307 +856 +278 

+412 +161 

—427 —638 

—540 —164 

—268 



64. Front Lift Truss — The solution of the front lift truss will be 
given in detail. The rear truss is analyzed in exactly the same way. 
Fig. 46 shows the outline of the truss in vertical projection, together with 
the loads at the panel points and the graphical solution for the stresses. 
The true plane of the truss is inclined to the vertical plane on account 
of the stagger, but a little thought will show that when the panel loads 
are vertical the stresses in the spars are correct when the vertical pro- 
jection of the truss is used. However, the stagger and slopes of wires 
will change the actual stresses in the struts and wires. The analytical 
and graphical methods are explained in Arts. 6 to 11 and 12 to 15, 
respectively. The analytical computations are given below. 
Stress in strut 1-1 = 92 lbs. compression (Method of Joints) 

Stress in spar 1-2, lower wing = Olbs. compression (Method of Joints) 

The stress in spar 1-2, upper wing, is found by the method of mo- 
ments, taking moments about station 2 of the lower wing. 

208 X 78.5 

Stress in spar 1-2, upper wing = = 292 lbs. compression. 

56.0 

The stress in wire 1 upper to 2 lower is found by the method of 
shears, using the principle that the vertical component of the stress in 
the wire equals the shear. This is true because the chords of the truss 
are parallel and horizontal. The shear equals the loads to the left of a 
section, or 92+116 = 208 lbs. 

96.5 

Projected stress in wire = X 208 = 358 lbs. tension. 

56 

81 



Wing Stress Analysis 



Art. 64 



0> 


A 

v> 

JO 




1 


8 
| 

■1 






<+. 











«3 , 






* 


? 




/"jo Vj 




V 


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It 


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5- 


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82 



Art. 65 Wing Spar Fittings 

The stress in strut 2 — 2 is found by applying %V = to joint 2 of 
lower wing. 

Stress in strut = 92 -j- 116 -f- 129 = 337 lbs. compression. 

The solution for stresses in members of panel 2 — 3 is similar to that 
just given for panel 1 — 2. Using the method of joints at station 3 of 
upper spar, the vertical component of stress in the cabane strut equals 

12 

96 lbs. This stress has a horizontal component of X 96 = 39 lbs., 

29.5 
and since the strut is in tension this component will produce 40 lbs. 
additional compression in the center section spar. By the method of 
joints at station 3 the stress in center section spar equals 886+39 = 925 
lbs. compression. 

The stresses given on the truss diagram in Fig. 46 are those ob- 
tained by the analytical computations. It will be seen that they are in 
very close agreement with those obtained by the graphical method. 

65. Wing Spar Fittings — The fittings for the front upper spar which 
transmit the stresses in the flying wires are shown in Fig. 47. These 
fittings are eccentric and very similar to the case taken up in Art. 38. 
The vertical bolts are the true reaction points for the spar, and the 
eccentricities of the struts and wires are given. The same type of fitting 
is used at all strut points on this airplane. 

66. Moments and Shears for Offset Strut Loads — Fig. 48 gives the 
loading on the upper front spar from the strut loads above. These loads 
are the stresses in the struts as determined in Art. 64. Case 1, given in 
Art. 172 for continuous beams applies to this case of loading. 

m^ 92X1.75 = +161 in. lbs. 

Apply the three-moment equation (Case 1) to spans 1 — 2 and 2 — 3, 
161 X 78.5 + 2 m 2 (78.5 + 56.75) + 56.75 m 3 — 
337X76.75 f 

I (78.5) 2 — (76.75) 

78.5 I 

12640 + 270.5 m 2 + 56.75 m 3 — +89520 
( 1 ) 270.5 m 2 + 56.75 m 3 — +76880 

Apply Case 1 to spans 2 — 3 and 3 — 4, noting that m 4 = m 3 because 
of symmetrv of wings about the center line of the airplane. 
56.75 m 2 -4- 2 m 3 (56.75 + 54.5) + 54.5 m 3 — 
56.75 m 2 + 277.0 m 3 — 

56.75 m a 

m 3 = 

277.0 

Substituting this value of m., in equation (1) : 

(56.75) 2 m 2 

270.5 m 2 == 76880 

277.0 

83 



Wing Stress Analysis 



Art. 66 




Frftina of Outer Sfrur Po/nt Q Fronf Spar. 




Hfh'ng at /nner Jfruf Po/nf© Fronf 3por 
© © @ 



SH 



Fig. 47. Wing Spar Wire Fittings. 



84 



Art. 67 



Moments and Shears for Eccentric Wire Pulls 





76880 

1 


+296.9 






258.9 






56.75 

1 


- 296.9 = - 


-61 


m 3 


277.0 


S-i = " 


-92 lbs. 






78.5 Sl 


+ 161 + 337 X 1-75 


= +297 




454 


-5.8 lbs. 




s i — 


78.5 




S_ 2 = - 


-337— (- 


-5.8) = - 


-331.2 lbs 


56.75 ^ 


2 + 296.9 


= —61 




s., = - 


357.9 


—6.3 







56. 


75 




















56.75 s_ 3 — 


61 


— 296.9 










S-g = +6.3 














s 3 = 














p<? # 




337* 






£ 


< 






© 






9 




@ 






% 


~% 


76$" 


vr 


— J. 
S6* 




S* 


* m 








73 h' 




.1. 




-9 


74* 






* 




+t 


w* 






*t 


S3* 



Fig. 48. Offset Strut Loads for Front Upper Spar 



67. Moments and Shears for Eccentric Wire -Pulls — The wire pulls 
and their eccentricities are shown in Fig. 49. The stresses in the wires 
were taken from Fig. 46. Case 5 given in Art. 172 applies to this con- 
dition of loading. 

M 1 = 358 X 1.10 = — 394 in. lbs. (Applied moment) 

M 2 = 762 X -84 = —640 in. lbs. (Applied moment) 

111! = 

2 m 2 (78.5 + 56.75)+56.75 m 3 — 394x78.5+2x640x56.75 

(1) 270.5 m 2 +56.75 m 3 — +103570 

56.75 m 2 +2 m a (56.75+54.5 ) +54.5 m 3 — 640x56.75 — 36320 

85 



Wing Stress Analysis 



Art. 68 



(2) 56.75 m 2 + 277.0 m, = +36320 

Eliminate m 9 between (1) and (2). Divide (1) by 270.5 and (2) 
by 56.75. 

(3) m 2 + .2098 m 3 = +382.9 

(4) m 2 + 4.882 m 8 = +640.0 



4.672 m 3 


= 257.1 


m 3 


= + 55.0 


m 2 + (.2098X55.0) = +382.9 


m 2 = +371.4 




78.5 Sl — 394--= 


+371.4 


+765.4 
78.5 


+9.7 


s. 2 = — 9.7 




56.75 s 2 + 371.4 


_640 = +55 


+323.6 
56.75 


+5.7 


s- 3 = — 5.7 




8,-0 








®J> 



^>S 



© 



^* 



762 



+ 9.7 



<?* 



ssi' 



4.0" 



S'7<5 



■ST 



-•5.7* 



Fig. 49. Eccentric Wire Pulls for Front Upper Spar 



68. Summary of Moments, Shears and Reactions for Front Upper 
Spar — Table VIII gives a summary which includes the effect of eccentric 
strut loads and wire pulls. 

86 



Art. 69 








Spar Properties 






TABLE VIII 








Distributed 


Eccentric 


Eccentric 


Result- 




Load 


Strut Load 


Wire Pull 


ant 


r x 


—115.8 lbs. 


— 97.8 lbs. 


+9.7 lbs. 


—203.9 


r 2 


—135.1 


—337.5 


—4.0 


—476.6 


r 3 


— 95.9 


+ 6.3 


—5.7 


— 95.3 


s -i 


— 46.1 


— 92.0 


0.0 


.—138.0 


S+i 


— 69.7 


— 5.8 


+9.7 


— 65.8 


S -2 


— 74.0 


—331.2 


—9.7 


—414.9 


s +2 


— 61.1 


— 6.3 


+5.7 


— 61.7 


S-3 


— 46.0 


+ 6.3 


—5.7 


— 45.4 


S +3 


— 49.9 


0.0 


0.0 


— 49.9 


m, 


+692 in. lbs. 


+ 161 in. lbs. 


in. lbs. 


+ 853 in. lbs. 


m 2 


+856 


+297 


4-371 


+ 1524 


m. 


+412 


— 61 


+ 55 


+ 406 


m x _ 2 


—638 






— 724 


m 2-3 


—164 






— 156 


m,,- 4 


—268 






— 274 



The maximum moment in each span is determined at the point of 
zero shear by the application of the formula expressing the relation 
between bending moments at different points in a span. In the follow- 
ing computations the bending moment at the point of zero shear is ex- 
pressed in terms of the bending moment and vertical shear just to the 
right of the adjacent support. It should be particularly noted that 
the bending moment just to the right of the support is equal to the bend- 
ing moment at the support changed by the moment of the eccentric 
wire pull. The necessary computations follow: 

Sj 2 wx- 

™ - ™ - - m, + s x x -f 



2 w 

65. 8 2 

— (853—394) = —724 in. lbs. 

2X1.83 

61. V~ 

m.,.., = (1524—640) = —156 in. lbs. 

2 X 1-83 

69. Spar Properties — The properties given in Table IX are com- 
puted by the exact method, but the approximate method is accurate 
enough. The sections of the spars in the spans are shown in Fig. 41. 

87 



a 









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Deflection of Front Upper Spar Art. 70 

70. Deflection of Front Upper Spar — The formula for deflection in 
a span with end moments and shears and a uniformly distributed load 
is given under Case 2 A in Art. 172. In using this formula it should be 
borne in mind that mi is the moment just to the right of the support at 
the outer end of the span. As an example of the application of this 
formula the deflection will be computed for span 1 — 2. The formula is: 

inj Sj W ~"| 

— + — (x + L)+-(x 2 + xL + L 2 ) | 
2 6 24 J 

Moment to left of support (1) =+853 in. lbs. 

Moment due to eccentric wire pull = — 394 



EIv = x (x— L) 



Moment to right of support (2) = +459 in. lbs. 

Shear to right of support (2) = — 65.8 lbs. 

w ===== 1.83 lbs. per in. L= 78.5 in. 

65.8 

When shear equals zero, x == = 36.0 in. 

1.83 

459 65.8 1.83 

EIv = 36.0(36.0— 78.5) (36.0+78.5H (36.0 2 

2 6 24 

+36.0X78.5+78.5 2 ) 

EIv = +370,300 

The value of / = 3.12 for the span may be taken from the table of 
properties in Art. 69 and £==1,600,000 lbs. per sq. in. Then v<= 

370,300 V 

0.0742 in. Maximum moment in span = m 1 ± 

2 w 

J i* . 
= — 723 in. lbs. The sign of is positive when 

2X1.83 lw 

s 1 is positive and negative when s x is negative. 

71. Drag — The subject of drag and resolution of forces has been 
taken up in Art. 48. As this analysis is for high incidence, a value of 8.0 
for the ratio of L/D will be taken with an angle of incidence of 12 
degs. Fig. 50 gives the resolution of forces and it will be noted that 
the drag component acts forward, or in a negative direction. 

72. Drag Components of Struts and Wires — Fig. 51 is a diagram of 
the struu. - + station 2 and the flying wires in panel 2 — 3. The loads 
shown in t%. "2 are taken from Table VII and are sufficiently exact 

89 



1,600,000X3. 


12 


65 


.8 2 


-459 





Wing Stress Analysis Art. 72 

for drag calculations. The components of stress may be obtained from 
the dimensions given in Figs. 51 and 53. The horizontal or drag com- 
ponent parallel to the chord = 11/56, or .196 times the vertical com- 
ponent of the stress in the member for both the struts and wires. A 
drag component acting forward is taken as negative, and one acting 
toward the rear as positive. 

DRAG COMPONENTS AT STATION 1 UPPER WING 

Member Vertical Component Horizontal Component 

Front Strut... 92 lbs. .196 X 92=— 18 lbs. 

Rear Strut. ... 39 lbs. .196 X 39 = — 8 lbs. 

Front Wire.... 92 + 116 = 208 lbs. .196X208 = +41 lbs. 

Rear Wire.... 39+ 49= 88 lbs. .196 X 88 = +17 lbs. 



Resultant Drag +32 lbs. 

The total drag, then, for the struts and wires is a concentrated load of 
32 lbs. acting towards the rear at the panel point of the drag truss. 

The calculation of the drag component of the stresses in the wires in 
panel 2 — 3 is made more difficult because of the divided wires from each 
of the upper strut points at station 2. It is necessary to make use of the 
theory of least work, which gives a simple solution. Fig. 51 shows these 
wires, and the first step is to determine the true lengths of the lift wires 
and the components of their stresses, assuming the vertical components 
to equal unity. As an example of this analysis, the computations of the 
stresses in wires C and G will be given in detail. 

Wire G Wire B 

(70.5) 2 = 4970.25 (70. 5) 2 = 4970.25 

(54.0) 2 = 2916.00 (54.0) 2 = 2916.00 

(9.25) 2 = 85.56 (36. 6) 2 = 1339.56 



(89.3) 2 = 7971.81 (96. 0) 2 = 9225.81 

Wire D Wire C 

(70. 5) 2 = 4970.25 (70. 5) 2 = 4970.25 

(56.0) 2 = 3136.00 (56.0) 2 = 3136.00 

(11.0) 2 = 121.00 (38. 4) 2 = 1474.56 



(90. 7) 2 = 8227 ,25 (97. 9) 2 = 9580.81 



90 



Art. 72 



Drag Components of Struts and Wires 




F*e cos 4°S?'30"~ .996G * E (for computations) 
O-R sin ^°S£'30'-.065/5 

£=W*f.006 

O=066F 



Fig. 50. Resolution of Forces at High Incidence Conditions 











^r-^i 










A\ /\ 


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i y\c V 






\ *"~""~ * 




T 


- e°* J 










— «* „ 





Fig. 51. Wiring in Inner Bay 



91 



Wing Stress Analysis 



Art. 72 




9e* T 39* 



Struts of 5 fa. 
5PS>R ^/T^CT/O/VJS FOG COMPUTZ/s/O D/P/F.T 



S J3<f 




/29^ \<?6> 



Struts of St a. ® 
5P/9/2 &£#CT/0/VS FO& COMPUT/NO D&/FT 

Fig. 52 




soz- 



" »L 



3' .?* 



^pj_ 



v55^" Chord 



Fig. 53. Wire in Outer and Center Section 



92 



Art. 72 Drag Components of Struts and Wires 







TABLE IXa 




Wire 
G 


Length 
89.3 in. 


Drag 
9.25 

89.3 


Comp. 
= —.104 


Stress/Vert. Comp. 
89.3 

=1.654 

54.0 


B 


96.0 in. 


36.6 
96.0 


- -.382 


96.0 

=1.779 

54.0 


C 


97.9 in. 


38.4 
97.9 


= +.392 


97.9 

=1.749 

56.0 


D 


90.7 in. 


11.0 
90.7 


= +.121 


90.7 

=1.620 

56.0 



The assumption will be made that the vertical load carried by the 
two wires G and C, or B and D is divided between them in such pro- 
portions that the work performed by the stresses is a minimum. 

For convenience let the vertical components of the stresses in the 
wires G and C be denoted by the letters G and C respectively. The 
total vertical component of the two wires may be taken from Table VII. 



From Art. 63. 






Ri 


(front lower spar) 


= 92.4 


Rx 


(front upper spar) 


= 115.8 


R 2 


(front upper spar) 


= 135.1 


R 2 


(front lower spar) 


= 128.8 



Total Vert. Comp. =472.1 lbs. 

Using the reaction given in Table VIII, R 2 — 476.6 lbs. 

There is very little difference for this particular airplane, but in some 
cases the reactions obtained by the method used for Table VIII will be 
more accurate. 

The stresses in the wires are: 1.654 G for wire G 

1.749 C for wire C 
G + C = 472 lbs. 

Let R denote the total work performed in stressing the wires. The 
expression for the work is given in Art. 116. 

(1.654G) 2 X 89.3 (1.749C) 2 X 97.9 

R = h 

2 AE 2 AE 

The two wires will be of the same size, so that the expression 2 AE 
is the same for both terms. 

93 



Wing Stress Analysis 



Art. 72 



R 



2AE 

C = 472 — G 

1 



244.3 G 2 + 299.9 C 2 



Let R 



2AE 



244.3 G 2 + 299.9 (222784 — 944 G + G< 



Differentiate R with respect to G and put the result equal to zero. 

dR 

= 488.6 G + 299.9 (—944 + 2 G) = 

dG 

1088.4 G = 299.9 X 944 

G = 259.9 

C = 472 259.9 = 212.1 

Stress in wire G= 1.654 X 259.9 = 430 lbs. 
Stress in wire C = 1.749 X 212.1 = 371 lbs. 
Drag component of wire G = — .104 X 430 = — 45 
Drag component of wire C = .392 X 371 = +145 



Resultant drag = 100 

In the same way, the stress in wire B is found to be 141 lbs., and in 
ire D 165 lbs. 

Drag component of wire B= — .382 X 141= — 54 lbs. 
Drag component of wire D= .121 X 165 = +20 lbs. 



Resultant drag = — 34 lbs. 

The drag component of the stress in the strut equals 11.0/56.0, or 
196 times the vertical component. 

Fig. 52 gives the reactions at the strut points at stations 1 and 2. 
Drag of front strut =.196 X (92+116+129)= —66 lbs. 
Drag of rear strut = .196 X (39 + 49 + 46)= —26 lbs. 



Total drag 



-92 lbs. 



SUMMARY OF DRAG COMPONENTS AT STATION 2 
Member Drag- 
Front strut — 66 lbs. 

Rear strut — 26 lbs. 

Wire G — 45 lbs. 

Wire B — 54 lbs. 

Wire C +145 lbs. 

Wire D +20 lbs. 



Resultant drag — 26 lbs. 

94 



Art. 73 



Air Load Drag 



73. Air Load Drag — The drag from the air load on the upper wing 
equals .085 times the load on the wing, and acts forward as shown in 
Fig. 50. It will be sufficiently accurate to assume that the drag load is 
concentrated at the panel points of the drag truss and that the panel 
points carry the drag load from half of the adjacent spans. In the case 
of the cantilever portion of the wing the entire drag will be assumed as 
concentrated at the panel point located over the outer struts. The drag 
load between the outer struts = — .085 (1.83 + .68) = — .213 lbs. per 
inch of run. 

74. Loads on Drag Truss — The loads at the panel points, as shown 
in Fig. 54, are made up of the drag from the air load, the thrusts of the 
struts, and the wire pulls. 

Load at Point (A) 

From cantilever = — .085 X 66.7 = — 5.7 
From panel A— B =—.213 X 19.6 = — 4.2 



From struts and wires = 
Total = 

Panel Point (B) 

Load = —.213 X 39.25 
Panel Point (C) 



—10 

+32 
-22 lbs. 



-8 lbs. 



From panel = — .213 > 
From struts and wires 



[39.25 -j-28.5 



= —7 
= —26 



Total 
Panel Point (D) 
Load = —.213 > 



= — 33 lbs. 



18.5+29.6 



-6 lbs. 



75. Struts — The dimensions of the struts are given in Table X. The 
properties of the sections were obtained with the aid of the following 
formulas: 

Area = .71 bd 

I (long axis) = .042 bd :: 

P (long axis) = .24 d 

These formulas were applied to the former U.S. A 
section for values of fineness ratio between 2.0 and 5.0. 
diameter was located at 40 per cent of the long axis 
Since the maximum diameter of the strut section of this airplane is at 
38 per cent of the long axis from the nose it will be accurate enough to 
use these formulas. The actual length of the front struts is 53^ in. and 



standard strut 
The maximum 
from the nose. 



of the rear struts 54 T / in. 



All struts are tapered. 
95 



Wing Stress Analysis 



Art. 75 



4: 

* < 


to 

A** \ 
/ 


it 


/6V i 


u " 
**> 


^ * / 

A 


*■ 

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-a ^ 

It 


Ab i 

/ * 


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\/ 




96 



Art. 76 



Stress Table 







TABLE X 






DIMENSIONS OF STRUTS 




Struts 

Front and Rear 
Outer 


Average 
Length 

54 in. 


Short Dia- Long Dia- 
meter "d" meter a b" 

IV. in. 4V, in. 


Fineness 
Ratio b/d 

3.78 


Front and Rear 
Intermediate. . 


54 in. 


l 15 / 32 in. 4 9 / 1(J in. 


3.12 



PROPERTIES OF STRUT SECTIONS 

Struts I Long Axis p Long Axis Area L/p 

Front and Rear— Outer 254 .27 3.40 200 

Front and Rear — Intermediate .608 .35 4.77 154 

As an example of strut calculations, the factor of safety for the 
front intermediate strut will be given. The strut will be assumed as 
pin-ended and the ultimate load computed from Euler's formula, using 
a value of E = 1,600,000 lbs. per sq. in. 

P tt 2 E tt 2 X 1,600,000 
Ultimate — = — = 665 lbs. per sq. in. 

A (L/p) 2 (154) 2 

Ultimate load = 665 X Area = 665 X 4.77 = 3180 lbs. 

This load is for a strut of uniform cross-section. These struts being 
tapered, the ultimate load equals .85 X 3180 = 2700 lbs. 

From Fig. 46 the vertical components of the stress in the strut 
equal 337 lbs. 

57.1 
Strut load — X 337 = 344 lbs. 



56.0 



The factor of safety = 



Ultimate load 2700 



Load 



= = 7.9 

344 



The front outer strut has a factor of safety of 12.1 

76. Stress Table — Table XI gives the properties of the spar sections 
at each of the strut points, and at the point of maximum bending 
moment in each span. Reference should be made at this point to Fig. 41 
in which sections of the front and rear spars are shown. The areas, 
section moduli, and radii of gyration are taken directly from Table IX. 

In the determination of the ratio of slenderness, L/p, the unsupported 
length for intermediate points is considered as the distance between strut 
points, and for inner strut points as two-tenths of the sum of the adjacent 
spans. 

The deflection at the point of zero shear in the outer span has already 
been calculated in Art. 70, and the same method may be employed in 
the determination of the deflection in the inner span. 

97 



Wing Stress Analysis Art. 77 

The moments in the next column are taken from Table VIII and 
represent the resultant moments due to the distributed load, eccentric 
strut loads, and eccentric wire pulls. Reference should be made at 
this point to Art. 51 in regard to the condition of maximum bending 
moments at strut points. It will be noted that the spar section at the 
strut points is solid with a vertical bolt hole deducted. At a point five 
or six inches from strut points routing begins. The strength of the spar 
at this point as well as at the strut point should be investigated, taking 
into consideration the change in bending moment, the modification of 
the spar section, and the reduction in the allowable unit stresses due to 
the effect of routing. 

The Moment Pd is due to the eccentric application of the direct 
stress and is obtained as the product of the resultant direct stress 
and the maximum deflection in the span under consideration. It 
should be noted that the eccentric moment varies as the square of the 
load factor since it is the product of two quantities, each of which varies 
directly as the load factor. The moment due to the distributed load 
and eccentric strut and wire loads varies directly as the load factor. 
Consequently it is necessary to multiply the eccentric moment by an 
assumed value of the load factor before computing the resultant moment. 
It may be necessary to make several trials before the assumed value 
of the load factor corresponds with the actual load factor. 

The ultimate allowable stress is determined as suggested in Art. 174 
for the case of laterally loaded columns. At the outer strut point the 
full compressive strength of the material is allowed in the calculation 
of the allowable fibre stress. In all other cases a reduction in the max- 
imum compressive stress is made to compensate for column action as 
recommended in the article referred to above. The modulus of rupture 
is here assumed as 8000 lbs. per sq. in. for the routed sections. The cor- 
rect value for the reduction of the modulus of rupture due to routing of 
the spar may be calculated as indicated in Art. 174. 

SPECIAL FORMS OF LIFT TRUSSES 

77. Clark Truss — The special features of this truss are clearly 
brought out in Fig. 55. There is only one lift truss, the lift wire attach- 
ing to the rear upper spar. The wire may be attached to the fuselage 
at the lower front spar as shown in Fig. 55, or at any other convenient 
point. The stress analysis of this truss is similar to the one given in 
this chapter, but it should be noted that the lift load at point A has to 
be carried down through strut AC and up through the strut CB before 
it can be transferred to the lift wire. In the same manner the loads at 
point C and D have to be transmitted to point B. 

For normal flight the rear upper spar is the only one carrying the lift 
compression, while the other three spars carry only bending moments, 
or moments and drift compression. This allows a saving in weight for 
the other spars, although increasing the size of the rear upper spar. In 
reversed, or upside down flights the front lower spar is the only one 

98 






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99 



Wing Stress Analysis Art. 78 

carrying lift compression. If the lift wire is carried forward still more 
than shown in the sketch it may be possible to relieve the compression 
in the rear upper spar by transferring the compression to the front upper 
spar. Trial computations will establish the correct location of the wire 
for the least weight of spars. 

The drift loads from the thrusts of the struts and the wire pull must 
be determined before solving for the stresses in the drift trusses. These 
trusses are best adapted for large stagger for the reason that when the 
lift truss is vertical the drift loads from the struts and wires are a min- 
imum. 

The chief advantage of a single lift truss is the lessening of the re- 
sistance due to the small number of wires. In a pursuit airplane the 
single truss has the great disadvantage that if one wire is shot away the 
wings will collapse. The upper front and lower rear spars may be dam- 
aged without injuring the lift truss. 

Fig. 60 illustrates the typical truss applied to multi-engined airplanes. 

78. S.P.A.D. Truss — This truss is arranged to support the spars in 
the center of their span by the auxiliary struts at station 2 in Fig. 56. 
The portion of the truss in action when in normal flight is shown by the 
full lines in Fig. 57. 

The loads W ^ and W 4 are equally divided between members 3 and 4. 
The vertical component of the stress in 3 is carried upward by the strut 
5 and then down through the wire 6 — 4. The horizontal component of 
the stress in 3 produces compression in the lower spar 9 — 10. The re- 
mainder of the stress analysis is similar to the one in the first part of 
this chapter. In landing, wire 11 is in action and wire 6 is inactive. 

This type is especially applicable in triplane construction, and in 
biplane construction where the span or loading is too great for an ece- 
nomical design of a one-bay airplane, yet not large enough to warrant 
a two-bay airplane. 

79. Thomas-Morse Truss — This form of truss, shown in Fig. 58, 
resembles somewhat that used on the S.P.A.D. airplane, but the lift 
and flying wires which cross at the inner interplane strut have no con- 
nection to the strut. It is evident that there is no tension on the wing 
hinges of the lower wing when in normal flight. The stress .analysis 
presents no difficulties. It may be noted that the lower wing spans 
may be shot away without endangering the upper wing or truss system. 

80. S.V.A. Truss — This airplane, Fig. 59, uses the familiar Warren 
truss in which the diagonals take either tension or compression. The 
diagonals are oval steel tubes with the ordinary crossed incidence wires. 
A drift wire (shown by the dotted line) leads from the rear upper spar 
to the nose of the fuselage. This type of truss permits an overhang 
on the upper wing without an increase in the strength of the spars. 
Rigidity and ease of assembly are two of the advantages of this form. 

81. Monoplane Wings — Partly because of a number of accidents 
due to faulty designing, the monoplane has not been as widely used as 
it should have been. There are several important advantages inherent 
in this type of wing construction. Aerodynamically, a monoplane wing 

100 



Art. 81 



Monoplane Wings 





C-D 
Fig. 55. Clark Truss 




Fig. 56. S.P.A.D Truss 




Fig. 57 



101 



Wing Stress Analysis 



Art. 81 




Fig. 58. Thomas-Morse Truss 




Fig. 59. S.V.A. Truss 




W 



K 




Fig. 60. Typical Wing Trusses of Multi-Engined Airplanes 



102 



Art. 82 Multi-Spar Construction 



is much more efficient than a biplane or triplane wing. From the stand- 
point of visibility a monoplane wing is excellent because there is no in- 
terference from a lower wing. If a pilot is so placed that his eye is on a 
level with the wing of a monoplane his blind angle due to the wings is 
very small. Even an externally braced monoplane cellule is much easier 
to assemble and true up than a biplane cellule, while an internally braced 
monoplane wing offers no rigging difficulties whatever. Alonoplanes 
braced by complicated trussing are entirely obsolete. If external bracing 
is employed it should consist of struts or wires from the front and rear 
spars to the bottom of the fuselage. If struts are used, the reactions 
from a negative air load are carried by them in compression, and if the 
wings are braced by wires against the direct air loads only, then the 
spars, which must in this case be continuous over the fuselage, acting 
as cantilevers carry in bending all negative loads. A wing structure 
should have a factor of 3.0 to 3.5 under reversed load. If a design is 
well adapted to a wing with continuous spars the wire type of bracing- 
is satisfactory as it offers a minimum of resistance. In comparing a 
wire braced wing with an internally braced one it should be noted that 
the former type when acting as a cantilever has only about 40 per cent 
of the load to carry that comes on the latter from the direct air load. 
For this reason one of the thin wing sections might be used on a wire 
braced wing if the greater efficiency of the thin wing is of much im- 
portance. For internally braced monoplanes the U.S.A. 27 wing section 
is excellent, combining as it does large spar depth, high lift and high 
efficiency. It will not prove as satisfactory on a strut braced cellule, 
as it is impossible to take full advantage of the spar depth afforded. 
Incidence wires in the plane of the struts on a braced wing should be 
omitted. If they are effective in carrying drag loads, they put such 
heavy secondary stresses in the struts that the latter fail at a low 
factor. All drag should be taken care of by the internal drag trussing. 
The use of both incidence wires and drag trussing is an example of the 
kind of an indeterminate structure that should always be avoided. 

82. Multi-Spar Construction — Wing construction of this type may 
be used to meet the specifications of wing strength and wing shape. 
Aerodynamic considerations may require a wing section so thin that it 
would be impracticable to use the ordinary two-spar construction be- 
cause of the uneconomical cross section of spar which strength require- 
ments would necessitate. 

Distribution of Load Between Spars 

In the stress analysis, the first problem that arises is that of the 
division among the spars of the distributed load. The problem will be 
discussed from two points of view: First, assuming that the deflections 
are equal and second, that the deflections may or may not be equal. In 
either case there are, as fixed conditions of the problem, the spar spacing, 
the spar stiffness denoted by the quantity EL and the wing loading. 
There are then three known or fixed conditions. With the first as- 

103 



Wing Stress Analysis Art. 82 

sumption accepted, i.e., that the deflections of the spars are all equal, 
there is a redundancy of limiting conditions. The problem can be solved 
by using two of the above known conditions and neglecting the third. 
First, answer the question with the spar spacing fixed and the spar 
stiffness known. Since the deflections of all the spars are equal, it can 

be stated that d = = d x = , etc., and since L and E 

EI EI X 

are equal, by writing W/I = W x /I x , etc., the proposition is developed 
that, no matter what the distribution of loading may be, the spars take 
a load in proportion to their moments of inertia. 

In the same manner, by holding to the two conditions of wing load- 
ing and spar stiffness, the proposition may be set forth that no matter 
what the spar spacing may be, the load is divided among the spars in 
proportion to their moments of inertia. 

Third, with the condition of loading and spar spacing as the two 
factors considered we may, by application of the "Three Moment Equa- 
tion," solve for the distribution of the load among the spars. In the case 
of a wing built with rigid compression ribs closely spaced, so that the 
wing construction approaches so-called "unit" construction, the assump- 
tion that the deflections of the spars are equal will closely approximate 
the truth. Hence, with such a type of construction the proposition has 
justification that no matter what the condition of loading may be, the 
distribution will be in proportion to the moments of inertia of the spars. 
In general it probably will not be economical to tie the spars together 
rigidly enough to obtain this unit construction, and in that case it is 
evident that manner of loading, spar spacing and spar stiffness are all 
vital factors, none of which may be neglected. The assumption that 
all the spars deflect equally may then be considered fallacious except 
in some special cases. 

One of the most important of these is the case of a cantilever or in- 
ternally braced wing. Here the deflection of the spars at all points along 
their length is in the same direction. By running a metal strap over the 
weaker or more heavily loaded trusses, such as the extreme front and 
rear trusses, and under the stronger, and generally less heavily loaded 
trusses in the deeper part of the wing section, or by otherwise suitably 
bracing between these trusses at several points, the deflection of all the 
trusses can be nearly equalized. Therefore, they will all carry loads 
closely proportional to their moments of inertia. In the construction of 
large monoplanes of the heavy bomber type this is an important con- 
sideration. 

For another case, make no assumption as to the relative deflection of 
the spars. Take for a specific example the condition as shown by Fig. 
61. Fig. 62 shows the method of distribution of the load between the 
spars. The wing is considered as being stiff enough to carry the load 
to the spars but not stiff enough to influence, through its rigidity, the 
distribution of the load to each spar. It is clear that spar spacing and 

104 



Art. 82 



Multi-Spar Construction 



condition of loading are the determining factors of the problem, and 
that the factor of spar stiffness does not affect the distribution under this 
assumption. In justification of this method of distribution of load 
between the spars the example in Fig. 63 is shown. Here a standard 
two-spar section is taken and the proportionate part of the load going 
into the front spar is computed by the method given in this paragraph, 
and the result compared with the value obtained by the standard 
method. 




Fig. 61 



-to*/ond*e 



10' 






\ /7* 


/3 






/4' 




' \ 


* 


3 


\\° 


1 


V 


1 lii* 


'M 


m\ 


9" 


1 liil 


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'O' ^ 


zo' 


>M 




e 


•1 p 
7* 


/<3 

















Computet/ on3 of £>fx?ct/ oa/s 



/ox ax re tyixjxjj =.sssw+.oz7W=jaeiv 



7ZXdi$sX4+^X3x'-f^*£ZZW-i-,os6 W*.E 76 W 

'J39W 
Fig. 62 



J 5X£X]fe 



105 



Wing Stress Analysis 



Art. 82 



. /9./0" 


S4" 










2V 






/6" 


t 2 - 


- /0 " ~ 




30" 




1 / 




' 




'i; 




to */ 


fo /<?f 


idZ 


j *?*? 










3 

4 W 







Computet /on of Load ts>/<ea/ by 3p/?e "7 

.SSSW -f ./07W^ .OJJPV =,69JW 

Dy Jt/?/vd/?£>d Method 

S<? X.354 - /g /o 

TO F&OA/T SP/)R = ?Q.g W= 697 1A/ 

so 
Fig. 63 

Division of Eccentric Moments and Shears Among the Spars 

These moments and shears in the upper wing are caused by eccentric 
strut thrusts or lift wire pulls. The points of application of these forces 
are on the center lines of spars 1 and 3. The forces are in about 
the same proportion to each other as are the stresses in the front and 
rear lift trusses. These forces are applied at the strut points, and since 
at these points the spars are very rigidly connected, they all act together 
as a unit. If the load were uniformly distributed over the wing section 
at this strut point it would be divided among the spars in propor- 
tion to their moments of inertia. But, since these forces are applied 
directly at the center lines of the outside spars, the division of the load 
will not be in proportion to the moments of inertia, but will approach 
that division where each outside spar takes all the load applied directly 
to it. The true division probably lies between these two outside limits 
and, considering the magnitude of the moments and shears caused by 
these eccentric thrusts and wire pulls, the error caused by arbitrarily 
taking the average between the results obtained by these two limiting- 
conceptions will not be outside the limits of precision of the complete 
analysis. 



Division of the Loads from the Lift Trusses 

The loads from the lift trusses are also applied at the center lines 
of the outside spars. These loads are to be resolved with one component 
in the plane of the spars. With stiff connections between the spars at 

106 



Art. 82 Multi-Spar Construction 

the strut points, and perhaps with the spars tied together at one or two 
places in mid span, the action is that of an eccentrically loaded column. 
The load on each spar may be computed by an adaptation of the formula 
L = A s (P/A -\- MXy/I)- Here L is the proportionate part of the load 
taken by each spar; A s is the area of the spar under consideration; P is 
the load whose effect is desired; A is the area of all the spars; / is the 
moment of inertia of all the spars about the center of gravity of all the 
spars; y is the distance of the spar under consideration from the center 
of gravity of all the spars, and M is the moment caused by the load P 
which is eccentric with respect to the center of gravity of all the spars. 

Solution of the Drag Truss System 

By assuming the drag truss to be pin- jointed and by neglecting the 
stresses brought into the spars because of the rigidity of the compression 
ribs, it is evident that there are no direct stresses from the drag truss in 
the intermediate spars. Fig. 64 represents a stress diagram for the drag 
truss computed under these assumptions. It is clear, however, that this 
solution can only approximate the truth and that it will not, except in 
a few cases, be of sufficient accuracy. By assuming stiff connections 
at the joints, the application of the theory of '"Least Work" to the prob- 
lem is justified. Fig. 65 shows the drag truss stress diagram with 
stresses computed by this method. For ease of computation, the as- 
sumption is made that the areas of the spars are equal. Referring to 
Fig. 65, and with particular reference to the stresses in the members cut 
by section AA, by taking moments about points 1 and 2 there can be 
obtained two independent equations involving three unknowns. The 
stress in the diagonal can be computed by the method of shear. The 
one equation of condition that is then required may be obtained by 
stating that the total work done by the members cut by the section A A 
will be a minimum. Fig. 65 shows the drag truss stresses in the spars 
as computed by this method. 

Comparison of the results as given by Fig. 64 with those as given 
by Fig. 65 shows that the simpler method of Fig. 64 is conservative and 
is close enough for design purposes, but that for an accurate stress 
analysis the method of Fig. 65 is probably better. 



107 



Wing Stress Analysis 



Art. 82 













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108 



CHAPTER IV 

MISCELLANEOUS DESIGN 

Ribs 

83. Plywood Web Type — For chord lengths greater than a certain 
value, the built-up truss rib is more economical than the plywood web 
rib. This value will vary for different types of wing sections, being less 
for the thicker types. For the sections represented by the R.A.F. 15, 
95 to 100 inches chord length is approximately the maximum. Ex- 
tensive testing and development work has shown that the rib section 
of the type shown in Fig. 66 is the best. For chords longer than 75 
inches, the capstrips should be increased in size. The web thickness, 
however, is probably satisfactory for all chord lengths between 60 and 
100 in. The function of the web in carrying shear and compression de- 
termines the direction of the face plies. Since the vertical members are 
in compression, and since only the material with the grain vertical is 
effective in compression, the best results are secured by having the face 
plies thin and placed with their grain vertical. The relatively thick core 



a* I (3ee F/q. 67) 

H J N. 




<6 Groin of r~~ace P//es l/erf/co/ 



Fig. 66. Cross-section of Standard Rib 




Jo eJ ao eo 
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Max/mum Depth of Zib */O0 



Fig. 67. Location of Cutouts in Terms of Maximum Rib Depth 



109 



Miscellaneous Design Art. 84 



used by spacing these plies far apart increases their moment of inertia. 
Numerous tests have demonstrated the superiority of an all-Spanish 
cedar web over a web with face plies of thin (1/70 to 1/100 in.) maple 
or birch veneer and a core of poplar or basswood, or a web in which 
both face plies and core are of some light wood such as pine, poplar, 
basswood, or spruce. The size, shape, and location of the web cutouts is 
another important matter. Fig. 67 gives what is perhaps nearly an ideal 
arrangement, but the presence of internal drift wires which must pass 
through the ribs may change this arrangement of the lightening holes 
somewhat, in order to secure clearance with all the ribs. The amount of 
material necessary in the web is dependent on the shear, hence the small 
cutouts near the spars. The ultimate strength of ribs of the type just 
described, for a chord length of 66 in., may be taken as 350 to 400 lbs. 
if the attachment of the rib to the spar is good. 

84. Attachment of Rib to Spar — Since many rib failures are due to 
improper methods of securing the ribs to the spar a consideration of 
the various methods of doing this is important. The difficulty is caused 
in transferring the shear from the web to the spar. Unless some positive 
means is provided for accomplishing this, the shear is carried by the 
capstrips, and if these are not properly fastened to the spar they are 
very liable to pull free from the web. Incidentally, the use of light, 
cement-coated nails to connect the capstrip and web will not prevent 
this. In general, nails should be used only to hold the members in place 
during gluing. A direct, and excellent method of securing the web to the 
spar is shown in Fig. 68. The use of triangular blocks is best adapted 
to box spar construction. The same idea can be carried out on routed 
spars with plywood blocks, as shown in Fig. 69. However, the general 
tendency of construction is to reduce to a minimum extra parts and 
weight. The use of a tenoned web illustrated in No. 2, Fig. 70 over- 
comes this disadvantage and accomplishes the same results as the blocks. 
There are two objections to this method; the fact that the ribs must be 
assembled on the spar, and that a shrinkage of the spar might cause 
serious splitting stresses in the latter if the tenon fitted closely. Yet the 
method is very satisfactory from the point of view of low weight, sim- 
plicity, and strength, and has gained considerable favor. Nos. 1 and 3 
of Fig. 70 illustrate two methods for preventing separation of the cap- 
strips and web. The rib is taped and glued in No. 1, and in No. 3 the 
capstrip is held by very light steel clips secured to the spar by 2y 2 inch 
screws. The clips are used only on the top of the spars, and unless the 
overhangs at the nose and tail are rather large, only on the inside edge, 
as is indicated in the photograph. However, in the case of machines 
which are to be stunted, a steel clip should be used on the top of the 
front spar on the outside as well as on the inside edge. If tape is em- 
ployed instead of steel clips, the ribs on such airplanes should be taped 
on the front side of the front spar also. For the reason that the screws 
have a tendency to split the spar flanges clips are not advised if the 
flange thickness is less than 3/4 in. In connection with these last two 
methods experiments have shown that the capstrips are strong enough 

110 



Art. 85 



Thick Sections 



in straight shear to develop the full strength of the rib if they can be 
prevented from pulling away from the web. The manner in which the 
wings are assembled is important in deciding on the means of attaching 
the rib to the spar. Complete assembly of the ribs before slipping on 
the spar is not recommended, because of the clearances that are neces- 
sary and the danger of breaking the glue joint between the web and 
capstrips. Also, fittings and blocks have to be put on the spar after the 
ribs are in position if this method is followed, a feature that is often 
undesirable. In summary it ma}' be said that methods employing little 
blocks or pieces of plywood are too laborious to be adapted to produc- 
tion. The added weight is also a serious item. The method involving 
tenoning of the web is difficult from the production standpoint. How- 
ever, if labor is not an important consideration the results obtained are 
excellent. The two methods recommended as satisfactory in every way 
are those using the tape and glue, or the steel clip. 

85. Thick Sections — In connection with a very thick wing section 
such as is found on the Fokker airplane it is suggested that several 
small vertical stiffeners of triangular section glued to the web between 
the spars add much to the stiffness of a thin veneer web, which relieved 
of its vertical compression can carry horizontal shear. 




Sect/ on /?/? 



Fig. 68. Attachment of Ribs 

to Spar with Triangular 

Blocks. 



Fig. 69. Attachment of Ribs 

to Spar with Plywood 

Blocks. 



Ill 



Miscellaneous Design 



Art. 86 




Fig. 70. Three Methods of Attaching Rib to Spar 



86. Split Caps trip — Instead of a single capstrip, grooved a 1/16 in. 
to receive the web, good results have been obtained in some designs 
with a double or split capstrip, as shown in Fig. 71. Because of the 
large gluing surface the likelihood of separation of the capstrips and web 
is reduced. If a very thin web is used it is practically impossible to 
nail vertically through the web, as is done with the one piece capstrip. 
But with the split capstrip horizontal nails can be used very effectively. 
The latter type, however, is the more difficult to assemble, partly be- 
cause of the extra piece, and partly because such a rib must be com- 
pletely assembled and then slipped on the spar, which, as previously sug- 
gested, requires undesirable clearances. 

87. Compression Ribs — In place of the usual double web compres- 
sion rib, the type shown in section in Fig. 72 is suggested as being more 
satisfactory. The size of the ribbed part of the web is calculated by 
treating it as a column with round ends as far as lateral bending is con- 
cerned. 

112 



Art. 



Truss Ribs 



L/ghf 3rads 




Fig. 71. Split Capstrip 



3prc/ce 



VTKZ^ZZZZZ 



'Capstr/ps some $/ze qj 
an ord/nory r/b 




r/pnye is 3+ro/ghf 



Fig. 72. Cruciform Compression Rib 



88. Truss Ribs — The Forest Products Laboratory recently com- 
pleted a series of tests on 15 ft. truss ribs of wood construction. Many 
different types of design were tried out, including Warren trusses and 
single and double Pratt trusses. The design found the most satisfactory 
as regards the strength-weight ratio, reliability, and simplicity of con- 
struction is the Warren truss, all-spruce rib illustrated in Fig. 73. With 
loading 2 of Fig. 76, which gives the most severe stresses, the average 
ultimate load for three ribs was 742 lbs.; with loading 1, 974 lbs. The 
average weight of these six ribs was 2.25 lbs. Failure seldom occurs in 
built-up ribs except at the joints, and the large majority of these fail- 
ures take place at one or more of the eight joints adjacent to the spars. 
Attention should therefore be directed toward securing strength at such 
points. The manner in which the joints of the rib in Fig. 73 are taped 
and glued provides considerable residual strength even should the gluing 
of the wood surfaces be imperfect. 

113 



Miscellaneous Design 



Art. 




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114 



Art. 88 Truss Ribs 

Fig. 74 is a modification of the construction discussed above. The 
principal difference is in the method of joining the web and chord mem- 
bers. The plywood gusset plates are effective. The chief objection to 
them is the labor entailed in screwing them to the chords and diagonals. 
This design is probably superior in reliability and ease of production to 
the rib in Fig. 73. Tests were made on three ribs of this design with 
162 in. chord. The average weight of the ribs was 31 oz. and their aver- 
age ultimate strength 1025 lbs. under a loading similar to loading 2, 
Fig. 76, but with the peak rounded off and with the load distributed 
between the upper and lower chord in the proportions of 1 to 2 re- 
spectively, instead of concentrated on the lower chord as is customary. 

The type of construction shown in Fig. 75 is a recent development 
in rib design. Its two main objects are elimination of weakness at joints 
and ease of production. The first object is attained by making the 
chords and web integral through the use of plywood; the second object 
by the elimination of complicated joints of the type used either on the 
Forest Products Laboratory or the Barling ribs. It will be noted that 
the diagonal reinforcing strips are uniform in section throughout their 
length and are cut square on the ends. But two, or at most three, sizes 
of diagonals are required, and two of these sizes can be the same as 
those of the chord members. A saving in weight can be effected by 
making the upper chord, which in normal flight is in tension except in 
the overhangs, smaller than the lower chord which is in compression. 
With an especially large rib it would be worth while to cut down the 
size of the lower chord to that of the upper on both overhangs, which 
are in tension for normal flight. The best type of plywood is one with 
spruce faces and poplar core, the plies being of equal thickness. The 
grain of the face plies is horizontal. For ribs from 100 to 120 in. long 
each ply should be about 1/32 in. thick; for lengths between 120 and 
150 in. 3/64 in. is a suitable thickness; for lengths from 150 to 180 in. 
each ply should be about 1/16 in. thick. Severe stresses are caused in 
the lower chord by the beam action of the chord between panel points. 
The stiffness of the chord can be readily increased by making the ply- 
wood leg of the chord section deeper. In calculating the stresses in a 
truss rib this secondary bending must be allowed for. Ribs of this 
plywood truss type are stiffer as a whole than ribs of other types, and 
their strength is less variable. 

Seven reinforced plywood truss ribs of a 106 in. chord length have 
been tested with loading 1, Fig. 76. Some of them were constructed 
with the spruce-poplar plywood recommended, and the balance with an 
all birch plywood of the same thickness. The first type weighed 15 to 
\S J / 2 oz. unvarnished, and carried an average ultimate load of 600 lbs. 
Those with birch plywood weighed 17 to 19 oz. unvarnished, and sus- 
tained an average maximum load of 800 lbs. The birch plywood ribs 
were appreciably stiffer than those with a spruce-poplar web. How- 

115 



Miscellaneous Design 



Art. 88 







116 



Art. 89 



Rib Loading 



ever, the latter were considerably lighter and developed all the strength 
that could be utilized efficiently. Beyond a certain point additional 
strength is of no value. All the diagonal reinforcing was 3/16 x 3/16 in. 
spruce, except the diagonals adjacent to the spars on the inside, which 
were 3/16 x 1/4 in. The upper chord reinforcement was 3/16 x 3/16 in. 
spruce; the lower chord, 3/16 x 1/4 in. Triangular or half round strips 
glued and nailed to the web of the rib and to the spar after the rib has 
been finally located on the spar make a good connection, especially with 
box spar construction. If an "I" section spar is employed the taping 
shown in Fig. 70 is advisable. 








fi/oh /nc/dence Lood/ng \ 

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(B) Loading /brusuo/ /nc/dence$ 




\£) Low /nc/dence lootf/no 
Fig. 76. Rib Loading for Static Test 



89. Rib Loading — For purposes of design and static testing the three 
types of loading given in Fig. 76 have been largely used. Loading 2 is 
the most severe for members between the spars, while the other loadings 
limit the design in the case of the nose and tail portions of the rib, espe- 
cially where the overhangs are unusually great. For static testing the 
total load is divided into equal loads spaced at 4 to 5 in., according to 

117 



Miscellaneous Design Art. 90 

the chord length. Where the type of sewing is such that all the load 
on the top surface of the aerofoil is transmitted to the lower chord of the 
rib, the entire load in a static test is applied to the lower chord. But if 
the fabric is sewed to the rib so that all the load on the top surface of 
the wing is carried by the upper chord, the load in static test is divided 
between the upper and lower chords, 50 — 70 per cent going on the upper 
chord. In design work provision should be made for reversed loading 
equal to half the direct loading. In calculating the ultimate load for a 
rib the factor required under sand test for the wing structure should be 
considered. 

90. Rib Spacing — Wing loading and maximum speed are both fac- 
tors in determining the spacing of ribs. For high powered pursuit air- 
planes a minimum spacing of 7 in. and a maximum of 10 in. are suitable. 
Within the slipstream and for perhaps a foot beyond, the minimum 
spacing should be employed. This may be gradually increased to the 
maximum up to a distance from the wing tip approximately equal to the 
chord length, beyond which the maximum may be used. For fast air- 
planes the maintenance of the true aerofoil section is important, which 
is one reason for the close rib spacing. In the case of two place air- 
planes with a high speed of 120 to 140 M.P.H., the maximum and min- 
imum limits of spacing may be taken as 9 and 15 in. With heavy bomb- 
ers the limits may be further increased to 15 and 20 in., depending on the 
size of the bomber. The maximum spacing of 20 in. should be suitable 
for airplanes with a gross weight of 20,000 to 25,000 lbs. 

For larger airplanes it is possible that an even greater spacing can 
be used. Rib spacing on airplanes of large size is a function rather of 
the strength of fabric than of rib design. With built-up truss ribs it is 
generally easier to design strong, somewhat heavy ribs suitable for a 
large rib spacing, than lighter ribs of less strength that would be used 
with a smaller spacing. It may be well to note that it is not so much 
the normal wing loading which determines the maximum load to be 
carried by the ribs as it is the dynamic wing load. For instance, with 
small, high-powered airplanes the maximum dynamic loading may be 
4 to 5 times the normal wing loading, while in the case of large bombers 
this dynamic loading would probably not exceed 1.75 times the normal 
loading. 

With all types of airplanes one false rib should be used between each 
two main ribs from the leading edge to the front spar. This should be 
of such a character that the true aerofoil section will be maintained. 

The use of thin veneer or plywood on the upper wing surface between 
the leading edge and the front spar, and preferably extending the entire 
length of the wing, is almost essential on airplanes of a speed greater 
than 120 M.P.H. For lower speed airplanes only the slipstream length 
need be covered. Mahogany plywood 3/64 in. thick has proved very 
satisfactory for this purpose. When no such reinforcing is used the rib 
spacing should be somewhat closer. 

118 






Art. 91 Streamline Section 



Struts 

For struts the forms in general use are the solid or hollow wood 
streamline type, and the elliptical or circular tubular type with which 
fairing is used. The properties of these various sections follow. 

91. Streamline Section — The streamline strut section which has been 
adopted by the U. S. Air Service is shown in Fig. 77. Its dimensions 
are given in Table XII. This section was originally supplied to the 
National Physical Laboratory by the Royal Aircraft Factory. It has 
the lowest air resistance of any section so far developed, and is very 
good structurally. 

TABLE XII 

DIMENSIONS OF STANDARD STREAMLINE SECTION 

Distance from Leading Edge Offsets from Center Line 

Decimal Part of "L" Decimal Part of "D" 



0.000 






0.0000 


0.005 






0.0638 


0.010 






0.1000 


0.020 






0.1437 


0.040 






0.2150 


0.070 






0.3000 


0.100 






0.3637 


0.150 






0.4350 


0.200 






0.4763 


0.250 






0.4900 


0.300 






0.5000 


0.350 






0.4950 


0.400 






0.4912 


0.500 






0.4675 


0.600 






0.4225 


0.700 






0.3612 


0.800 






0.2850 


0.900 






0.1887 


0.950 






0.1237 


0.980 






0.0762 


0.990 






0.0500 


1.000 






0.0000 




Radius 


of Nose = .12 D 




2 properties of the solid 


section may be accuratelv expressed in 


of the length a 


nd diameter. 






Area, 


A = .730 LD 








About long axis About short axis 


Moment of Inertia, I 


.0432 LD 3 


.0432 L 3 D 


Radius of Gyration, p 


.243 D 


.243 L 






119 





Miscellaneous Design 



Art. 92 




Fig. 77. Standard Strut Section 



The properties of the hollow streamline section of this contour may 
be expressed in terms of the ratio of the mean length to the mean dia- 
meter. The mean length equals the outside length minus the thickness 
of the tube, and the mean diameter equals the maximum outside dia- 
meter minus the thickness. The ratio of these two distances may be 
defined as the mean fineness ratio to distinguish it from the true fine- 
ness ratio of the outside dimensions. The formulas apply only for 
mean fineness ratios between 2.0 and 4.5 and for a maximum ratio of 
wall thickness to outside diameter of 1/10. For greater wall thickness 
it would be more accurate to lay out the strut section to a large scale, 
and by scaling the ordinates to the inside of the section, compute the 
properties of the hollow portion which may be deducted from those of 
a solid section. However, for both streamline and elliptical sections the 
values of their properties, as given by the formulas in Art. 91-93 in 
which are used constants taken from the charts, become less than their 
true values as the ratio of t/D becomes greater. It is, therefore, always 
safe to use the formulas. 

Let L = mean length. D = mean diameter. R = mean fineness 
ratio = L/D. Ki = moment of inertia constant. Kp = radius of 
gyration constant. Ka = area constant (also mean perimeter constant) 

I = KitD 3 A = K A td p = K P D 

Ka= 1.875R + .992 

Ki = .290 R + . 054 (about long axis) 

.060 

Kp = .394 (about long axis) 

R 

These constants may be read off the diagram of Fig. 78. 

92. Hollow Circular Tubing — The exact formulas for the properties 
of a hollow circular section are very simple and easy to apply. 

Let d = outside diameter, d x = inside diameter, and t = thickness. 

120 



Art. 92 



Hollow Circular Tubing 



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121 



Miscellaneous Design Art. 93 



D = mean diameter = 

2 
-(d^-d, 2 ) 

Area, A = = ?r Dt 

4 

7r(d^— d x 4 ) 
Moment of Inertia, I = 



Section Modulus, S 



Radius of Gyration, p 



64 
-(dW,*) 

32d 



Vd 2 + d x J 



When the thickness, t, does not exceed 1/10 of the outside diameter, 
d, the following approximate formulas may be used with no appreciable 
error : 

A = tt Dt (exact) I = .394 D 3 t P = .354 D 

93. Hollow Elliptical Tubing — In Fig. 79 is given a set of curves for 
the property constants for various mean fineness ratios. The mean fine- 
ness ratio is the same as that for hollow streamline sections. These con- 
stants give results of sufficient accuracy when the ratio of thickness to 
outside diameter is not greater than 1/10. 

Area = Ka Dt 

Moment of Inertia = Ki Dt 

Radius of Gyration = KpD 

When the ratio of thickness to outside diameter of hollow sections is 
as large as 1/10 the error of the various approximate formulas used 
above is about 1 per cent on the safe side. 

94. Design of Struts — Interplane struts should be calculated as 
round ended columns. Euler's formula, P = ir EI/L 2 , should be used 
for computing the strength of struts when the L/p for the strut equals 
or exceeds these values: fir 90, spruce 100, steel 125. E for spruce 
or fir may be taken as 1,600,000 lbs. per sq. in.; for steel 29,000,000 lbs. 
per sq. in. When the values of L/p fall below the values that have been 
given, then the design of wood or fir struts should be based on the col- 
umn curve in Fig. 181, and the design of steel struts on the column 
curve in Fig. 25. Center section struts will usually be included in this 
class of short struts. In calculating the strength of faired, hollow, or 
elliptical steel struts, it is customary to neglect the increase in strength 
afforded by the fairing. This fairing may be considered as offsetting 
the effect of any slight eccentricity in the strut due to the fact that it is 
liable to be slightly bent. In Euler's formula the strut is assumed to be 
perfectly straight. 

122 



Art. 94 



Design of Struts 



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5^? 



Miscellaneous Design Art. 94 

When steel and wood are used in combination in a strut section, the 
strength of the composite section can be accurately determined by the 
formulas given in Art. 25. Should it be desirable to include the fairing 
in computing the strength of a faired strut, it may be done in the same 
way. 

Instead of using directly the values obtained from the wing analysis 
for the loads on the struts there is a tendency to somewhat increase 
these stresses to allow for the secondary stresses put into the struts by 
the incidence wires. Incidence wires may be stressed on account of: (1) 
Initial tension; (2) their tendency to equalize the loads carried by the 
front and rear lift trusses; (3) their tendency to equalize the deflections 
of the upper and lower drift trusses. It is suggested that an increase 
of about 15 per cent be made in the original strut stresses. 

Regarding the practice of laminating struts, it may be said that other 
things being equal, the strength of a strut is reduced by the use of lami- 
nations. Considered from a military standpoint it is probable that a 
laminated strut is less likely to be shattered or splintered by bullets than 
a solid strut. Laminating cross or spiral grained material will not im- 
prove its strength. In general, small or medium sized struts should not 
be laminated if material is available for making them from solid stock. 
On the other hand, in the case of large struts it is usually more con- 
venient to build them in two parts. Very often large struts are routed 
out in the center to lighten them. 

The walls of hollow wood struts should not be less than 3/8 in. 
thick, and in the case of very large struts 1/2 in. thick, if danger from 
warping is to be avoided. In this connection it may be well to suggest 
that in the case of hollow steel struts the use of material thinner than 
.035 in. is not advisable. For struts of streamline section the gage 
should be .049 in. or heavier for diameters greater than about \y% in. 
For circular or elliptical struts the gage should be increased from .035 
in. to .049 in. at a diameter of about 1*4 in. 

Best results are obtained from wood struts when the grain in the 
cross-section of the strut makes an angle between 45° and 90° with the 
long axis of the section. Any tendency toward warping or twisting is 
minimized. In a strut built up of two halves the grain should run as 
indicated in Fig. 80. 

It is interesting to note that an Euler strut may have its section 
largely reduced by piercing with holes without its strength being im- 
paired at all, providing that the direct stress on the net section is fairly 
well below the elastic limit of the material in compression. Both test 
and theory bear out this fact. For this reason the strut section shown 
in Fig. 80 is recommended for military airplanes subjected to severe 
fire. Tests have demonstrated that a bullet piercing such a strut at 
any angle will leave a clean hole. The plywood prevents splitting and 
the canvas prevents the splintering of the wood which would occur with 
an ordinary spruce strut. 

Probably the best combination of wood and steel for a strut is shown 
in Fig. 81. The difficulty with such a section is in securing a proper 

124 



Art. 95 Combined Bending 



S Spocer Tube 

6e 5pon/3h cedar 
p/i/tvood 





^--Conyos , o/ued to strut 

~~~Stee/ -5?rop 

Pig. 80. Built-up Strut Fig. 81. Combination Wood and 

Steel Strut 

distribution of stress between the two materials. Because of the shear 
between the wood and steel the latter tends to slip. An effective means 
of preventing this action is the use of rivets, 3/16 to 1/4 in. in diameter, 
and spacer tubes. These require closer spacing near the ends of the 
strut than at the center. The steel strips should form part of the fitting 
at each end of the strut. Struts of this type are particularly suited to a 
braced monoplane cellule in which they will be in compression only 
occasionally. 

Hollow streamline steel struts are of comparatively recent develop- 
ment. They have several important advantages, one of which is their 
suitability for production. Their equivalent weight, on all but very 
slow airplanes, is less than that of solid wood struts, and their reliability 
is much greater. Since the strength of slender struts is dependent on 
the modulus of elasticity, and not on the yield point or tensile strength 
of the material, ordinary mild steel may be used. 

Elliptical tubing with fairing is seldom used for interplane struts, 
but more often for center section struts or for chassis struts. Balsa 
wood has been found satisfactory for fairing, as it is very light. So- 
called "portal" or "picture frame'' struts built up of laminated wood are 
becoming well known. They present certain decided advantages, as 
elimination of incidence wires, and simplicity in rigging, but are con- 
siderably heavier than the usual type. For center section struts the 
"N" type, of either laminated wood or steel is frequently employed. 
Both "N" and portal struts are more expensive to manufacture than 
single struts. In neither type should the struts be tapered, since this 
adds greatly to the difficulty of production. 

95. Combined Bending — When a strut is subjected to bending mo- 
ments about both axes it may be necessary to determine what point on 
the strut section is most severely stressed. The curves given in Fig. 82 
will give the co-ordinates of such a point in terms of the short diameter 
when the ratio of the two moments is known. These curves are pre- 
pared for a streamline strut of a fineness ratio of 3.5, and an ellipse 
of a fineness ratio of 1.7. 

96. Taper of Struts — A curve for the standard taper of interplane 
struts is given in Fig. 83. It is derived from analytical considerations, 
and tests on struts with this taper have shown that the strength is nearly 
uniform throughout their length. The strength of a tapered strut may 
be taken as 85 per cent that of a straight strut of the same maximum 

125 



Miscellaneous Design 



Art. 96 



diameter and length. The ratio of the weights and volumes of tapered 
and untapered struts of the same maximum diameter and length is .774; 
the corresponding ratio for projected areas is .871. 



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126 



Art. 97 



Resistance of Struts 



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97. Resistance of Struts — The resistance of a strut with the standard 
streamline section and a fineness ratio between 3.5 and 4.5 is given by 
the formula 

R = .000204 LDV 2 
where R = resistance (lbs.) 

L = length of strut (ft.) 
D = diameter of strut (ft.) 
V = velocity (M.P.H.) 

This resistance should be multiplied by the VD correction factor 
taken from the diagram, Fig. 84. As an example, find the resistance of 
a \ l /\ in. diameter strut 6 ft. long, when the velocity is 90 miles per hour. 



1.2. 



R = .000204 X 6 X 



12 



X 90X90 =1.03 lbs. 



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The correction factor for VD = 9.4 is 
sistance = 1.03 X 1-03 — 1.06 lbs. 

When a tapered strut is used the resistance should be computed for 
a straight strut with a diameter equal to the maximum diameter of the 
tapered one, and this resistance corrected by the ratio of the projected 
areas. For a standard taper this ratio of areas is .871. It is impossible 
to keep exactly similar sections in a tapered strut and as a result the 
air resistance of the strut is somewhat increased. 

127 



Miscellaneous Design. 



Art. 98 



££3/3T/?/VCE Or ST&UT3 . 
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V - ve/ocity m.ph. 

l-Mox thickness ofstrui 
in feet 




98. Fineness Ratio — As the velocity increases the resistance is less 
for struts with higher ratios. The following fineness ratios are sug- 
gested for use on airplanes with various maximum speeds. 



Maximum Speed 

Over 175 M.P.H. ... 
Over 140 M.P.H. . . . 
Over 100 M.P.H. . . . 
Less than 100 M.P.H. 



Fineness Ratio 

4.5 
4.0 

3.5 
3.0 



99. Equivalent W 'eight— -When comparing two different struts it is 
sometimes of value to compare their equivalent weights, which are ob- 
tained by combining the resistance of a strut with its actual weight.. The 
resistance of a strut at any velocity is reduced to an equivalent weight 
by multiplying the resistance of the strut in pounds by the correct value 
of L/D (lift/drag of the airplane). A common value for this L/D ratio 
for high speed is 3.5 to 4.5. Equivalent weight = resistance X L/D + 
weight in pounds. 

For experimental design work, the weight and resistance of a strut 
may be expressed in terms of some variable, such as a dimension, which 
is to be determined. Then, by differentiating the expression for the 
equivalent weight and putting the result equal to zero, the value of the 
unknown quantity which will give the minimum equivalent weight can 
be computed. For instance, suppose that a comparison is desired be- 
tween wood and steel struts, or that the most efficient wall thickness for 
hollow struts' is to be found. Both of these things can be obtained by 
applying the principle of equivalent weight. 

100. Center Section Struts— -These struts are most severely stressed 
under a reversed loading or possibly in rolling. They should be designed 
to carry the compressions occurring in reversed flight. In the ordinary 
type of center section, with four struts and diagonal incidence and cross 
wires, the struts should be strong enough to develop the strength of the 
cross wires, that is to carry a load equal to the vertical component of 

128 



Art. 100 



Center Section Struts 



the stress in the cross wires when these are stressed to their ultimate 
strength. The struts should be designed as pin-ended columns using 
the column curves of Figs. 181 and 25 for wood and steel. Faired, round 
or elliptical tubes are often used for center section struts. 

The "N" type of strut, shown in Fig. 85, of either steel tubing or 
laminated wood, is quite frequently used. It is a rigid type of construc- 
tion and easier to assemble than the separate struts and wires. The 
diagonal member should be designed to carry in compression all the 
drift loads coming on the center section. 

Still another kind of center section construction is indicated in the 
line drawings, Fig. 86, which are front and side views of what may be 
termed the tripod type. With this construction the necessity for diag- 
onal cross wires is removed, thus giving the pilot a clear view ahead. 

Since the calculation of the stresses in such a system is somewhat 
difficult a solution will be made for the system shown in Fig. 86. The 
same general methods are used that are explained in Chapter V under 
the solution for the stresses in chassis struts and wires. In Table XIII 
are given the components of the cabane struts and wires expressed as 
decimal parts of the lengths of the members. 

TABLE XIII 
COMPONENTS OF CABANE STRUTS AND WIRES 

(Expressed as decimal parts of the lengths of the members) 
Front Cabane 
Component Drag Wire Lift Wire Strut 

Vertical —.756 —.967 +.765 

Horizontal (drift) —.497 +.094 

Horizontal (Perpen. to Prop, axis) . . +.429 +.237 

Rear Cabane 
Component Front Strut Lift Wire 

Vertical +.588 —.957 

Horizontal (drift) ... +.300 —.107 

Horizontal (Perpen. to Prop, axis) . . — .748 +.251 

Note — For vertical components, + is upward. 

For horizontal, drift components, + is backward. 

For horizontal components perpendicular to propeller axis, 

+ is toward center line of airplane. 

^ H/nge 

fr 

I 
I 

Fig. 85. N Type of Center Section Strut 
129 



—.486 
—.420 



Rear Strut 
+.428 
—.668 
—.613 




Miscellaneous Design 



Art. 100 











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System 



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Fig. 86. Tripod System of Center Section Struts 

130 



Art. 100 Center Section Struts 



The conditions of equilibrium are applied to the components of all 
the forces and loads applied at a point and resolved along three axes at 
right angles. The three equations so obtained are solved simultaneously. 
Each component of the stresses is expressed as a percentage of the total 
stress in the member, using the values given in Table XIII. As an 
example, the equations for the reversed flight condition are given below: 



Front System 

1285 lbs. = vertical load 
942 lbs. = drift load 
2V =— 1285 + .765 Cs = 
2D = + 942 — .486 Cs — D = 
2H = + H — .420 Cs = 
Cs = 1680 lbs. = compression in front strut 
D = — 126 lbs. = drift force necessary for equilibrium 
H = +705 lbs. = tension in spar necessary for equilibrium 



Rear System 

782 lbs. = vertical load. 

942 + 126 = 1068 lbs. drift load. 
2V= —782 + .428 Cr + .588 Cf = 
2D= + 942 + 126 — .668 Cr + .300 Cf = 
2H= + H — .613 Cr — .748 Cf == 
Cf = 125 lbs. = compression in front strut. 
Cr = 1656 lbs. = compression in rear strut. 
H — +1108 lbs. = tension in spar necessary for equilibrium. 

The cabane struts and wires should be calculated for the high and 
low incidence, diving, and reversed flight loading conditions. They 
should be further investigated for a lateral load perpendicular to the 
propeller axis. In the design under consideration this latter loading 
proved a limiting case only for the strut of the front system which, as 
designed, can carry a lateral load of 1100 lbs. plus a loading under the 
high incidence condition corresponding to a factor of 6.5. 

The front system is not stable within itself because, under several 
conditions, the drag wire tends to be put in compression. In those cases 
this wire was assumed out of action, and the horizontal forces necessary 
for equilibrium were applied to the system. For reversed loading none 
of the wires in either the front or rear systems f.re stressed. 

131 



Miscellaneous 


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d 



Miscellaneous Design Art. 103 



103. Built-up or Laminated Beams — Of the very large number of 
types of built-up spars developed, the type possessing the greatest advan- 
tages for spars of ordinary depth is that shown in Fig. 87. For the 
flanges straight-grained, clear material should be used. But in the web 
either cross or spiral-grained material will give entirely satisfactory re- 
sults. The principal features in favor of this spar may be summarized 
briefly: i. The small size of the material that can be utilized. 
ii. The possibility of employing defective stock in the web. in. Sim- 
plicity and low production cost. iv. Reliability and strength of the 
construction due to the broad gluing areas and opportunity of applying 
heavy, direct pressures in the gluing operation, v. Minimizing of 
shrinkage stresses which with some types of built-up sections are very 
severe. 



, . /or . i 
I i — S/ooe f:/0 r 1 



-6 \ m /06 ,| 



i 



s 



W£e Spj./C£ Fiahoc Spl/ce 

Three P/ece Qu/lt-up Spars 



Dotted //net 3/>or* 
spar before routing 



Fig. 87. Cross-Section and Splice for Built-up Beam 



The spar described in the preceding paragraph is, of course, intended 
as a substitute merely for the usual "I" section, one-piece spar that is 
used in the so-called "thin" wing sections such as the U.S.A. 15, 16, or 
the R.A.F. 15. For this purpose it is much superior to any kind of 
laminated construction in which a number of thin laminations of wood 
are glued together, generally with the grain of all the plies parallel. All 
laminated construction is heavy because of the weight of the large 
amount of glue required. It is also relatively expensive to manufacture. 
The laminating and gluing of wood do not increase its strength, hence 
the modulus of rupture of a laminated section is no greater than if the 
same material were in one piece. For the same reason it is not possible 
to utilize inferior material in such a section and get the strength that 
would be obtained from normal material. 

Recently several "intermediate" or "thick" wing sections have been 
developed, especially for cantilever construction. Because of the depth 
of these wings the customary one-piece spar section cannot be used. Up 
to the present, the "box" spar with solid flanges of spruce or ash and 
webs of plywood has been generally used, with not entirely satisfactory 
results. Such a member has a modulus of rupture intermediate between 

134 



Art. 103 



Built-up or Laminated Beams 



that for a one-piece beam, which with spruce varies from 10,300 lbs. 
per sq. in. to about 7,500 lbs. per sq. in. depending on the routing, and 
the ultimate compressive strength, which for spruce is 5,500 lbs. per 
sq. in. It may be taken as between 6,500 and 7,000 lbs. per sq. in. when 
the flanges are spruce. If the spar is deeper than 8 or 9 in. either the 
plywood must be fairly thick, and consequently heavy, or else small ver- 
tical stiffeners, as shown in sketch III, Fig. 88, must be used, in which 
case the plywood can be made thinner. Best results for box spars have 
been secured with the grain of the plies at an angle of 45 degs. to the 
horizontal and with all the plies of equal thickness. It is not advisable 
to have a shoulder on the sides of the flanges to protect the top of the 
plywood, since unless the flange is very deep the amount of gluing sur- 
face between the webs and the flanges is seriously curtailed. The ply- 
wood should, in general, run up nearly flush with the outside of the 
flanges. For box spars the thickness of the webs will vary from 3/32 
to 3/16 in. depending on whether the plywood is of hard wood, such as 
birch, or of spruce and poplar, and depending also on the size of the spar. 
For greater depths than 9 or 10 in. the construction shown in Sketch 
I, Fig. 88, is well adapted. Such a beam is light for its strength because 
all the material is efficiently used, much more so than in the ordinary 



^^~ 





/?-/? 




n 



sd 



3-3 
Fig. 88. Built-up Spar of Box Type 



135 



Miscellaneous Design Art. 103 



box construction. It is believed that greatest strength will be secured bv 
having the grain of the face plies run parallel to one set of diagonals, and 
that of the core, parallel to the other diagonals. As this type of beam 
is in process of development this point has not been definitely deter- 
mined. The core thickness should be 50 per cent of the total thickness 
of the plywood. Somewhat greater thickness of plywood can be eco- 
nomically used in this construction than in box beams; 1/8 to 1/4 in., 
depending on the character of the wood used and the spar size, is a suit- 
able range. It will be noted that the reinforcing diagonals are of uni- 
form section throughout their length and that the ends are square. In 
calculating the stresses in the members of this type of spar, the spar 
should be considered as a truss. The vertical component of the stress 
in a diagonal equals the shear at that point. The direct compression 
or tension in the flanges equals the bending moment at any section di- 
vided by the distance between the centers of gravity of the flanges. The 
diagonals can be assumed as fixed ended in one direction and as half 
fixed in the other. To allow for this difference in fixity the diagonal re- 
inforcing is rectangular rather than square in section. The flanges can 
be taken as fixed ended columns with an unsupported length equal to 
the distance between panel points. The reduction in stress due to col- 
umn action will usually be so small as to be negligible, and the full ulti- 
mate compressive strength of the material may be used. The strength 
of such a truss beam will be less variable than that of a box spar. Truss 
beams are particularly applicable to internally braced wings. The 
flange size can be readily varied according to the stress. It should 
also be noted that where, as in cantilever construction, the direction of 
the bending moment does not change, the lower flange, which is in ten- 
sion, can be made smaller than the upper or compression flange. The 
spar must, however, be calculated for a reversed load equal to a factor 
of about 40 per cent that for direct load. In the case of a large airplane 
in which a braced cellule and an "intermediate" wing section are em- 
ployed the bays could be made so large that it would be worth while to 
reduce the size of the upper flange near the center of the bay when the 
bending moment is negative, and to reduce the size of the lower flange 
at the strut points at which the moments are positive. The flange which 
is in compression must of course be increased in section where the ten- 
sion flange is decreased. 

What is considered the best type of steel or duralumin spar, espe- 
cially for deep beams, is shown in sketch II, Fig. 88. The flanges are of 
heavier gage material than the webs since the web stresses are small 
compared with the flange stresses. Local crinkling or buckling in the 
compression flange is prevented by having the flange curved. Metal 
ribs can be easily attached to the spar by riveting through the inner edges 
of each flange. All the material in a truss spar of this type is effective. 
The remarks in the last paragraph about decreasing the size of the 
tension flange are also applicable to metal spars. A change in spar 
section can be easily effected by riveting together flanges of different 
gages. 

136 



Art. 104 



Splices in Spars and Longerons 






104. Splices in Spars and Longerons — The Forest Products Labor- 
atory of the U. S. Department of Agriculture has conducted a series of 
tests to determine the best method of splicing spars (Fig. 89). The 
results are assembled in a report entitled "Project L-228-3, Tests of 
Airplane Wing Beam Splices." The conclusions set forth in that report 
are quoted as follows: 



Plan 



^ ^hardis/bo dDo tve/s f 



El£VS*T/ON 

Two Piece: La^in^ted DE#r*i 




ELEi/ar/OA/ 



Fig. 89. Spar Splices 



1. "Of the four different kinds of scarfs tested the plain sloped scarf 
is the best." 

2. "In solid beams and parts of built-up beams in which the height 
is less than the width, the plane of the scarf should be perpen- 
dicular to the plane of bending; that is, the scarf should be 
diagonally across the side." 

"In beams, or parts of beams, in which the height is greater than 
the width, the plane of the scarf should be vertical; that is, the 
scarf should be diagonally across the top. In beams with height 
equal to width either type of splices may be used." 

3. "The tests show that, for splices in spruce beams with high-grade 
glue and good gluing, a slope of 1 to 10 is sufficient to develop 
practically the full strength of the section." 

4. "Roughening the scarf does not increase the efficiency of a splice." 

5. "Dowels or bolts do not materially affect the efficiency of a well- 
made splice; they do not come into action to any extent until the 
glue fails. Now if the glue fails in the splice in a one-piece (un- 
laminated) beam, dowels or bolts ordinarily will not prevent an 
immediate collapse of the beam unless the joint is subjected to a 
moment less than 20 per cent of the maximum moment which the 
unspliced beam will withstand. In the two-piece laminated beam, 
however, with a splice in one lamination, dowels or bolts may 
keep up the strength enough to prevent collapse if the moment at 
the joint is not more than 65 per cent of the maximum moment 
which the unspliced beam will withstand. Although results of 
tests are slightly more favorable with bolts than with dowels, 
dowels are, as a rule, preferable in practice." 

137 



Miscellaneous Design Art. 105 



Design of Fittings 

Most fittings used in airplane construction depend upon bolts 
or pins to transfer stresses from one member to another . It is 
essential that the bolts have sufficient strength in shear, bearing, and 
also in tension, if necessary. These considerations are easily taken 
into account. Such fittings as clevises, lugs, eye-bolts, turnbuckles, 
etc., have been standardized so that the proper sizes for given loads 
may be chosen from tables without stress calculations. A few of 
the general features of the design of these fittings will be taken up to 
serve as guides for the design of special cases. 

105. Bolts and Pins — Table XVI gives the cross-sectional area of 
several sizes of pins or bolts. In the case of bolts in tension the net area 
should be taken at the root of the threaded portion. 

TABLE XVI 
AREAS OF BOLTS AND PINS 

Diameter Area 



1/16 in. 


.0031 sq. 


in. 


3/32 in. 


.0069 sq. 


in. 


1/8 in. 


.0123 sq. 


in. 


3/16 in. 


.0276 sq. 


in. 


1/4 in. 


.0491 sq. 


in. 


5/16 in. 


.0767 sq. 


in. 


3/8 in. 


.1105 sq. 


in. 



Where pins or bolts are used the stress is often transferred through 
a clevis. This puts the pin in double shear, i.e. there are two areas re- 
sisting the shearing forces. For example, determine the size of pin 
required to transmit a stress of 600 lbs. in double shear, assuming that 
the material of the pin has an ultimate shearing strength of 50,000 lbs. 
per sq. in. and that a factor of safety of 4.0 is desired. 

4 X 600 

Required area of pin = = .0240 sq .in. 

2 X 50,000 

The nearest size pin to this has a 3/16 in. diameter and an area of 
.0276 sq. in. 

A method of dealing with bolts which bear on wood will be derived. 
First assume that the fittings are not wrapped around the spar but are 
similar to those shown in Fig. 90. 

Let P = component of pull in a direction parallel to the neutral 
axis of the spar, 
a = distance from neutral axis to line of action of the com- 
ponent, P. 
d = depth of spar, 
b = diameter of bolt. 

138 



Art. 106 



Lugs and Eye-Bolts 




Fig. 90 



The bearing pressure of the bolts against the wood is composed of a 
uniform pressure due to the load P and a uniformly varying pressure 
due to the moment Pa. Let / c = unit pressure due to load P, and / b = 
unit pressure due to moment Pa. When there are two bolts, as in Fig. 90, 
assume the load and moment to be equally divided between the bolts. 

With one bolt: P 6 Pa 

f c = and f b = 

bd bd 2 

P 6 Pa P 

fc + fb = \ — 

bd bd 2 bd 



6a 
d 



The sum of / c and f b shall not exceed 5500 lbs. per sq. in. for spruce. 

When the bolts pass through a metal sleeve or fitting which entirely 
surrounds the spar there can be no exact method of solution, but the 
assumption may be made that the bolts carry 25 per cent of the 
moment in bearing on the wood, the fitting transferring the remaining 
75 per cent. 

The component of the pull at right angles to the spar is taken en- 
tirely by the bolts, and sufficient area of fitting under the bolt heads 
must be supplied in order to distribute the bearing pressure. 

If the pull-on a bolt, which bears on wood, has a component that 
makes an angle with the grain of the wood then care must be exercised 
in securing the wood against splitting. 

106. Lugs and Eye-bolts — As the result of a series of tests on lugs 
the following procedure in their design has been suggested. Given the 
desired maximum strength: 

I. The pin size is chosen in conjunction with the lug thickness, so 
that the maximum bearing stress shall not be more than twice the 
tensile strength of the material. The pin must be large enough to 
carry safely the shearing stress. Pin sizes for lugs of different 
capacities are given in Table XXXIII in the appendix. 

II. The diameter of the head should be made sufficiently large 
that computed stress in direct tension on the net section through 
the eye shall, at the maximum desired capacity, not exceed the 

139 



Miscellaneous Design 



Art. 107 



tensile strength of the material. It is necessary, however, to 
offset the center of the radius for the outer part of the head to 
care for extra stresses in thehead due to bending. The amount 
of this offset for each lug capacity is given in Table 33. 

III. The width of the shank should at least be sufficient to give 
shank area enough to stand the maximum desired load in tension. 

IV. The hole in the lug should be equal to the diameter of the 
pin, with limits of — .000 in. and -(-.010 in. 

Eye-bolts may be used when the angle of pull is not greater than 45° 
with the axis of the bolt. Beyond that limit the strength of the shank 
decreases very rapidly. When the eye-bolts bear directly against wood 
instead of a fitting they bend and crush the wood in front of them. This 
bending raises the apparent strength of the bolts by decreasing the angle 
of pull. Where the eye-bolt passes through wood enclosed in or cov- 
ered by a metal fitting the ability of the fitting to distribute the stress 
in the bolt over a sufficient area of wood is of importance. The larger 
sizes are unsuited for use at critical sections or in important members, 
because of the amount of material which has to be removed in the bolt 
hole, with the resultant weakening of the member. 

In lugs and eye-bolts of high strength steel it is well to offset the 
center of the pin hole by 1/32 of an inch from the center of the head in 
order to provide additional metal in front of the pin. With high strength 
steel the metal does not yield as readily as with ordinary steel, causing 
the pin to bear more nearly on a point than on an extended surface. 

107. Horizontal vs. Vertical Bolts — It is evident that a bolt passing 
vertically through a spar affects the moment of inertia of the spar much 
more than a horizontal bolt at or near the neutral axis. Horizontal bolt 
holes near the top and bottom of spars are as detrimental to the strength 
of the spars as are the vertical holes. Due allowance for bolt holes 
should be made, wherever they occur, and the size of hole should be 
taken as 1/32 in. to 1/16 in. larger than the diameter of the bolt in 
order to allow for the injury to the fibres. One good method which is 



=* 



55-^i- 



t 



<ft fr S 



Pl/P/V 



Elei/sit/on 



Fig. 91 
140 



Art. 108 



Pin Plates 



employed with vertical bolts is illustrated in Fig. 91. The vertical bolts 
are placed on the sides of the spar and are held in position by blocks 
which are glued to the spar. Two horizontal bolts at the neutral axis 
also assist in holding the blocks to the spar. Such an arrangement as 
this overcomes the objection to vertical bolt holes in the spar itself. 

108. Pin Plates — In many cases it is possible to save weight in the 
metal fittings by using thinner material and brazing a pin plate to the 
fitting at the bolt hole in order to reduce the bearing stress. An example 
will illustrate the method. Referring to Fig. 92, which represents a por- 
tion of a fitting carrying a pull of 840 lbs., assume that a factor of safety 
of 5.0 is desired, and that the steel has the following ultimate strengths: 




tVroppec/ Fitt~/n<p 



CW9 



Fig. 92. Lug with Pin Plate 

Ultimate shearing strength = 45,000 lbs. per sq. in. 
Ultimate tensile strength = 60,000 lbs. per sq. in. 
Ultimate bearing strength = 100,000 lbs. per sq. in. 

The ultimate pull on fitting = 840 X 5 = 4200 lbs. per sq. in. 

4200 

Tensile stress on body of fitting = =59,700 lbs. per sq. in. 

3/32X3/4 
The 1/4 in. pin or bolt is in double shear. 

4200 

Shearing stress on pin = = 43,000 lbs. per sq. in. 

2 X .049 
First assume that there is no pin plate. The bearing stress will 
4200 

equal = 179,000 lbs. per sq. in. This value is much in ex- 

1/4 X 3/32 
cess of the ultimate bearing strength, although the remainder of the 
fitting is satisfactory. Let t = required thickness to give a bearing 

4200 
stress of 100,000 lbs. per sq. in.; t =- - = .1680 in. The 



1/4x100,000 



141 



Miscellaneous Design 



Art. 109 



thickness required is (.1680 — .0938) =-.0742 in. in excess of the thick- 
ness of the fitting. The pin plate should be at least .0742 in. in thickness 
and in this case a value of 3/32 in. will be chosen. The final bearing 
4200 

stress equals = 89600 lbs. per sq. in. 

1/4X3/16 
on bolt: The component of the stress parallel to the beam 
2940 lbs. 

2940 
= 82400 lbs. per sq. in. 



Bearing 
axis is 4200 X .70 

Bearing stress on bolt 



_ .25 (.0930 + .049) 
109. Wing Fitting — Fig. 93 is a wing-spar fitting developed by the 
Airplane Design Section. This is essentially a box fitting and illustrates 
very clearly how eccentricities of struts and wires may be avoided. It 
will be noted that the lines of action of the struts and wires intersect at 
a common point on the center line of the spar. The channel across the 
top of the spar acts as a bearing plate to transfer a portion of the strut 



load directlv into the wood 




Fig. 93. Wing Spar Fittings Without Eccentricities 

142 



CHAPTER V 

AIRPLANE CHASSIS 

110. General Considerations — Until recently comparatively little 
had been done on the accurate stress analysis of chassis. Consequently, 
there has been less opportunity to corroborate by static test the results 
of the present methods of chassis analysis than is the case with the wing 
structure. Some uncertainty also exists in regard to the proper factors 
of safety and conditions of loading for various types of airplanes. How- 
ever, the methods of analysis given in this chapter are based entirely 
on the principles of mechanics and it is believed will give a reasonably 
exact solution for the stresses produced by known external loads, 
and therefore form a suitable basis for design. Further experimental 
work may be largely directed toward obtaining a better knowledge of 
the loadings that will reproduce actual conditions of service, and more 
correct methods for computing the ultimate strength of struts subjected 
to combined bending and compression. 

111. Conditions of Loading — The external loads to which a chassis 
may be subjected can be resolved into vertical forces, horizontal forces 
acting in a direction parallel to the propeller axis of the airplane, which 
may be termed backward or forward thrusts, and horizontal forces per- 
pendicular to the propeller axis, which may be termed side thrusts. 
Each of these forces, applied initially to the wheels, is transmitted to the 
axle, and from the latter into the chassis structure itself, either by means 
of elastic cords or by direct bearing of the axle on the axle guides. It is 
customary to assume that the backward, forward, and side thrusts are 
each a certain percentage of the vertical load. For airplanes of different 
types and sizes these percentages should vary. In present analysis the 
back thrust is taken equal to .5 of the vertical load, the forward thrust .2, 
and the side thrust .25. 

Below are given different loadings which should simulate nearly all 
the conditions that will be found in service. In each case the propeller 
axis is considered horizontal. 

Case 1. Vertical load in level landing plus horizontal back thrust 
sufficient to make resultant of load pass through center of 
gravity of airplane. This component will usually be be- 
tween .16 and .25 depending on the type of airplane. 

Case 2. One wheel landing with vertical load. 

Case 3. Case 1 plus horizontal back thrust equal to .5 of the ver- 
tical load. 

Case 4. Case 2 plus horizontal back thrust equal to .5 of the ver- 
tical load. 

Case 5. Case 1 plus horizontal forward thrust equal to .2 of the ver- 
tical load. 

Case 6. Case 2 plus horizontal forward thrust equal to .2 of the ver- 
tical load. 

Case 7. Case 1 plus side thrust equal to .25 of the vertical load. 

Case 8. Case 2 plus side thrust equal to .25 of the vertical load. 

143 



Airplane Chassis Art. 112 

Case 9. Case 1 plus both backward thrust equal to .5, and side 
thrust equal' to .25 of the vertical load. 

Case 10. Case 2 plus both backward thrust equal to .5, and side 
thrust, in the direction of low wing, equal to .25 of the ver- 
tical load. 

To calculate the dynamic factor for each of these cases would be 
impracticable and also unnecessary. The type of airplane should de- 
termine not only the factor needed, but also the cases that should be 
calculated. For example, a training airplane is liable to be subjected to 
treatment both more severe and varied than would an airplane in the 
hands of a skilled pilot. The principle to be followed in selecting the 
cases to be considered is that sufficient cases should be investigated 
to stress severely each member of the chassis. Case 1 does not pro- 
duce such stresses, but, because most chassis tests have been made with 
this loading, the dynamic factors so obtained have a certain comparative 
value. Case 4 is a limiting condition for the rear strut and usually for 
the diagonal wires. Case 6, which is intended to reproduce the condi- 
tions of a tail-low landing, is apt to be a limiting case for the front strut. 
These three cases should always be computed, and in addition it is ad- 
visable to investigate Case 7 which may prove to be the limiting case for 
the diagonal wires. 

112. Characteristics of the Chassis Structure — There are several fea- 
tures which are important in the analysis and design of a chassis: I. The 
method of attachment of the spreader tubes to the struts or the guide 
plate: If the tubes are pinned they will carry compression only, and 
may be very light, while if rigidly connected they will also serve to trans- 
mit bending moment, thereby somewhat relieving the struts. It is 
necessary in this case to strengthen the spreader tubes. With a rigid 
connection, the fixity of the struts in column action is increased. II. The 
character of the fitting securing the chassis struts to the longerons: In 
some instances this is in the form of a socket receiving the strut end 
which is bolted through and rigidly held. A connection of this type 
will practically fix the upper end of the strut. Aside from largely in- 
creasing its column strength, such a connection will affect the distribu- 
tion of the moments in the struts, or between the struts and spreader 
tubes. When this fitting is hinged, the pin may be either in the plane 
of the struts, that is parallel to the propeller axis, or it may be perpen- 
dicular to this direction. The first type is the more common. As the 
pin is parallel to the long axis of the strut the latter must be computed 
as a column hinged at one end at least. Moments perpendicular to the 
plane of the struts must become zero at the pin, but in determining the 
distribution of moments in this plane the struts are assumed to be fixed. 
On the other hand, when the pin is perpendicular to the plane of the 
struts, moments in this plane must reduce to zero at the pin, and the strut 
may be taken as fixed in determining the distribution of moments per- 
pendicular to the plane of the struts. Although the struts are un- 
doubtedly partially fixed at their upper ends it is hardly safe to count on 

144 



Art. 113 



Concerning the Chassis Used as an Example 



much increase in column strength from this cause. III. The bracing 
wires: Whether diagonal bracing wires are in the planes of both the front 
and rear struts, or in the plane of one set of struts only, affects the 
methods of analysis and also the distribution of the stresses. IV. The ec- 
centricities of the members in a chassis: One of the marks of a good de- 
sign is the absence of such eccentricities, or their reduction to an amount 
where they affect the stresses but slightly. Eccentricities which produce 
moments perpendicular to the plane of the struts are usually the most 
serious. Because in a level landing the struts and spreader tubes are in 
action, while in a one-wheel landing the struts and wires are the mem- 
bers that are stressed, it is possible to have an eccentricity in one case 
and none in another. Fig. 94 illustrates this. Slight eccentricities that 
cause moments in the plane of the struts are difficult to avoid and do 
not cause large stresses. As shown in Fig. 95 both struts and wires may 
be eccentric. V. The material used in the struts: Wood and steel are 
the materials usually employed. It is in the calculation of the allowable 
stresses and in solutions involving the method of least work that the 
nature of the material affects the analvsis. 




a 5preoder Tt/be 5 X Yfo 






-+h 



Fig. 94 



113. Concerning the Chassis Used as an Example. 

I. The following conditions of loading are investigated: cases 

I. 4, 6 and 7. 

II. The weight of the fully loaded airplane is 1650 lbs. 

III. The design of this chassis is such that in front elevation 
there is no eccentricity of struts, wires or spreader tubes. In side 
elevation, a slight eccentricity of struts and axle is present, as 
well as an eccentricity of the wire. Fig. 95 gives line drawings 
of the chassis members, showing the eccentricities. 

IV. The struts in their own plane are assumed to be fixed at 
the ends. 



145 



Airplane Chassis 




Art. 113 




kj 



1 



k 

I 

I 









-T3 

SI 






w 






§ 

s 



146 



Art. 114 Kinds of Stresses To Be Considered 

V. In determining the distribution of the moments in a vertical 
plane perpendicular to the plane of the struts, it is assumed that 
the connection between the struts and the vertical guide plate 
and between the spreader tubes and this plate is rigid. 

VI. The struts are hinged at their upper end with pins running 
in the direction of the propeller axis. 

VII. In the actual chassis there are wires in the plane of the 
front struts only, but to illustrate the method of solution when 
wires are present in the plane of the rear struts too, the case of 
a one-wheel landing will be worked out for this latter condition, 
in addition to the solution for the original chassis. 

114. Kinds of Stresses to be Considered — As previously suggested, 
there are two types of stresses: those due to direct column loads, and 
those caused by bending moments. The methods for obtaining the 
column loads and bending moments are different and will be presented 
separately. Bending moments are due to two causes: fixed eccentricities, 
and eccentricities produced by the rise of the axle in its slot. 

115. Calculation of Direct Stresses — The first step in the analysis is 
to compute the theoretical length of each of the members in the chassis, 
and the length of each of their projections on three axes at right angles 
to each other, one axis vertical, the other two parallel and perpendicular 
to the propeller axis respectively. The length of each of the projections 
should be expressed as a decimal part of the length of the member. Then 
the values of the components, along the three axes, of the unknown stress 
in a member may be expressed by the corresponding decimal parts of 
the unknown stress. Since it is necessary to consider the direction of 
these components the conventions shown in Fig. 96 are followed. In 
Table XVII the struts are assumed to be in compression, the wires and 
spreader tubes in tension. 



1 


> Prop /lx/3 




t- 



2 

Fig. 96. Convention for Signs of Forces 



147 



Airplane Chassis Art. 115 

TABLE XVII 

COMPONENTS OF STRESSES IN MEMBERS 

Vertical Drift Side 

Rear Strut —.558 C r —.786 C r +.265 C r 

Front Strut — .774 C f +.555 C f +.306 C f 

Spreader Tubes 000 T s .000 T\ -1.000 T s 

Front Wire +.566 T f —.394 T f —.723 T f 

Rear Wire (assumed) . . +.446 T r +.625 T r —.640 T r 

C r = stress in rear strut 

C f = stress in front strut 

T s = stress in spreader tubes 

T f = stress in front wire 

T r = stress in rear wire. 

Solution by Statics — Whenever the structure is statically determin- 
ate, that is where not more than three members carrying stress meet at 
a point, the principles of statics are employed. The conditions of 
equilibrium — namely, that the algebraic sum of the components along 
each of three axes at right angles, of any system of forces which is in 
equilibrium must equal zero — are applied to the components of the 
forces in question. Where but one set of wires is used the direct stresses 
can be obtained by statics for all types of loading. But, in the special 
case that is being given, where there are wires in the planes of both 
front and rear struts, the principle of least work must be called upon to 
obtain the value of the fourth unknown after the other three have been 
computed in terms of the fourth by the method of statics. 

The external loads are considered separately, so that the stress in 
the different members produced by each type of load can be obtained. 
Then the total stress caused in any member by any combination of 
loads is merely the sum of the separate stresses produced by the single 
loads. For convenience, each external load is taken as unity. Hence, 
to determine the actual stress produced in a member by any given load, 
it is necessary only to multiply the unit stresses by the amount of the 
actual load. 

Following are the equations of equilibrium for each different type 
of load considered, and the stresses in the members for a load of unity. 
The case of a vertical load in a one-wheel landing where two sets of 
wires are in action is completely worked out by the method of simul- 
taneous equations. The full solution for the same case with but one set 
of wires in action is given to illustrate the use of determinants. This 
latter process, for statically determinate cases, is simple and somewhat 
more rapid than the method of simultaneous equations. 

Case 1. Vertical load of unity: level landing. In this case it is 
assumed that the diagonal wires are not in action. 

5V= — .558 C r —.774 C f +1.000 = 
2D=— .786C r +.555C f =0 

2S = +.265 C r +.306 C f —1.000 T s = 

148 



Art. 115 



Calculation of Direct Stresses 



C r = -f-.605 compression 
C f = +.855 compression 
T s = +.421 tension 



Case 2. Vertical load of unity: one-wheel landing 
that the spreader tubes do not act in this case. 



It is assumed 



SV=— .558 C r 
2D= —.786 C r 
%S = +.265 C r 



.774 C f +.566 T f +1.000 = 
.555 C f — .394 T f =0 

.306 C f — .723 T f =0 



C f = +1.470 compression 
C r = + .613 compression 
T f = + .847 tension 



*In this part of the analysis the sign given the stress does not 
denote tension or compression, but merely indicates whether the 
direction of the stresses assumed in writing the equations was 



right or wrong. 



Solution of Case 2 bv method of dete 



|— .558 —.774 +.566 
I— .786 +.555 —.394 
I +.265 +.306 —.723 



C, 



rminants. 

—1.000 -^.774 +.566 

.000 +.555 —.394 

.000 +.306 —.723 



+ .459 



C r 



.610 



+.401— .121 = 4-.280 



I— .558 —1.000 +.566 

.459 Cf |— .786 .000 —.394 

(+.265 .000 —.723 



C f = +1.464 



+.104 + .569 = .673 



.459 T f = 


—.558 —.774 - 
—.786 +.555 
+.265 +.306 


-1.000 
.000 
.000 


+.241 + .147 = 
Tf = +.845 


= .338 


Explanation of Determinants 
a x — b y + c z = d 
a' x + b'v — c' z = d' 
— a"x + b"y + c"z — d" 




149 







Airplane Chassis 



Art. 115 




a 


— b 


c 1 




a 


d 


c 1 


a' 


b' 


-c'l 


y = 


a' 


d' 


-c'i 


-a" 


b" 


c"| 




— a" 


d" 


c"| 


a 


— b 


c 1 




a 


— b 


d 1 


a' 


b' 


-c'l 


z = 


a' 


b' 


d'| 


-a" 


b" 


c"l 




—a" 


b" 


d"| 



[ (ab'c" + a'b"c — a"c'b) — (— cb-'a" — c'b"a — c"a'b) ] X= 
(db'c" + d'b"c + d"c'b) — <(cb'd" — c'b"d — c"d'b) 



The determinant, which is the coefficient of x, y and z, is invariable. 
In multiplying, due attention should be paid to the signs of each term. 

The few brief explanations of mathematical work given in this sec- 
tion and elsewhere are inserted merely for the convenience of the reader 
to save possible reference to tables of integrals or to a handbook. 



Side thrust of unity perpendicular to axis. 

C r — .774C + .566T, =0 






.55 



Case 3 
2V 

1D = — .786 C r + .555 C f — .394 T f 
2S = +.265 C r + .306 C £ — .723 T f + 1.000 



Cf = +1.457 compression 
C r = -j- .022 compression 
T f = +2.010 tension 

Case 4. Thrust of unity parallel to axis. In this case it is assumed 
that the diagonal wires are not in action. 

2V = —.558 C r — .774 C f = 

2D= — .786 C r + .555 C £ + 1.000 = 
2S = +.265 C r + .306 C f — 1.000 T s = 

Cf = +.842 compression with back thrust 
C f = — .607 tension with back thrust 
T s = +.038 tension with both back thrust and 
forward thrust 

Case 5. Vertical load; one-wheel landing (both sets of wires) It 
is assumed that the spreader tubes do not act in this case. 

150 



Art. 115 Calculation of Direct Stresses 



(1 ) 2V = —.558 C r — .774 C f 4- .556 T f + .446 T r + 1.000 = 

(2) 2D = —.786 C r + .555 C f — .394 T f + .625 T r = 

(3 ) 2 S = +.265 C r + .306 C f — .723 T f — .640 T r = 

Divide (1), (2) and (3) by coefficients of C r 

(4) — 1.000 C r — 1.386 C f + 1.014 T f + .799 T r + 1.793 = 

(5) — 1.000 C r + .706 C f — .502 T f + .795 T r =0 

(6) +1.000 C r + 1.154 C f — 2.728 T f — 2414 T r =0 

Subtract (4) and (5) 

(7) — 2.092 C f + 1.516 T f + .004 T r + 1.793 =0 

Add (5) and (6) 

(8) +1.860C f — 3.230 T r — 1.619 T r —0 

Divide (7) and (8) bv coefficients of C f 



(9) — 1.000 C f + .725 T f 4- .002 T r + .858 


=0 


(10) +1.000 C r — 1.736 T f — . 870 T r 


=0 


Add (9) and (10) 





— 1.01 ITr — . 868 T r + . 858 = 
T f = — .858 T r + .848 
C t =— .620 T r + 1.473 
C r = +.789T r + .610 

Solution for T r by method of least work. (See Art. 1 16) 

C r 2 L CYL T, 2 L T r 2 L (+.789 T r + .610) 2 48.1 

WtotaI= 1 1 1 = h 

2AE 2AE 2AE 2AE 2 X -228 X 29,000,000 

(.620 T r + 1.473) 2 38.1 (— .858 T r + .848) 2 57.7 

-I + 

2 X .228 X 29,000,000 2 X -0313 X 29,000,000 
T r 2 X63.8 



2 X -0313 X 29,000,000 
d W 2 X -789 (+.789 T r + .610) 48.1 



+ 



dT r 2X.228E 

-2 X .620 (— .620 T r + 1.473) 38.1 
+ 



2 X-228E 
—2 X .858 (—.858 T r + .848) 57.7 2 T r X 63.8 



= 



2 X-0313E 2X.0313E 

151 



Airplane Chassis Art. 116 

The lengths of members used are the actual, not theoretical lengths. 
Solving this equation: 

T r = + .388 
T f -+ .516 
C f = +1.232 
C r = + .916 
The assumptions made that in the different cases the diagonal wires 
or spreader tubes were or were not in action are justified by investiga- 
tions in which no such assumptions were made. The stresses in the 
members that were assumed out of action were shown to be negligible. 
116. The Method of Least Work — As in the example just given, 
there are certain cases for which a solution is impossible by the methods 
of statics, and another condition or principle must be employed. This is 
known as the method of least work and may be stated as follows: when 
a structure is acted upon by a balanced system of external forces, the 
stresses will be so distributed that the total internal work done in the 
various members due to their deformation will be a minimum. The 
procedure is first to express the work done in each member by the ten- 
sion, compression, or bending to which that member is subjected, in 
terms of the unknown stress or bending moment. By taking the sum 
of all these expressions for the work in the individual members, an 
equation is obtained for the total work in the structure in terms of one 
or more unknown stresses. In order to determine at what value of the 
unknown stress or moment the total work will be a minimum, this equa- 
tion must be differentiated with respect to the unknown quantity, and 
the resulting equation put equal to zero. A solution for the stress or 
moment will give its correct value. In case there is more than one un- 
known stress the process of differentiation must be performed as many 
times as there are unknown stresses, the total work being partially dif- 
ferentiated with respect to each unknown in turn. All the unknowns 
except the one in question are treated as constants during the differ- 
entiations. The equations so obtained are solved simultaneously in the 
usual manner. 

In a member carrying direct stress only, the expression for the in- 
ternal work is W = PX 3/2 = P 2 L/2 AE. ' ' 

In a member subjected to bending stresses only, the expression for 
the internal work is, W = \m 2 dx/2 EI 

P = Total end load, tension or compression, expressed in terms 

of the unknown stress. 
L = Length of the member. 
A = Cross-sectional area of the member. 
E = Modulus of elasticity of the material. 
8 = Total deformation of the member. 
I = Moment of inertia of the section. 

m= Bending moment, expressed in terms of the unknown quan- 
tity and X. 
W= w x + w 2 + w 3 + . . . . = total internal work in structure. 

152 



Art. 117 Distribution of Moments 

The origin in each case is taken at one end of the member for which 
the work expression is being written. 

The process of differentiation is purely mechanical and not at all 
difficult. With direct stresses, practically the only types of expression 
to be differentiated are in the form W '== aP 2 or W = a{b — cP) 2 . The 
differentials of these expressions are dW/dP = 2aP and dW/dP = 
— lac (b — cP), respectively. The quantities a, b and c are known con- 
stants. In the second type, if the sign is + both signs in the differ- 
ential are -|— 

This method of work is far more accurate than any arbitrary way of 
proportioning stresses, since it is based only on mathematics and on the 
properties of the members, and contains no assumptions. Once under- 
stood, its application is very simple. 

For further explanation of the method, reference can be made to 
Fuller and Johnson's "Applied Mechanics," Vol. II, p. 242, and to 
Spofford's "Theory of Structures," pp. 359-367. 

117. Distribution of Moments — In discussing the distribution of the 
different moments between the members of a chassis, the case of the 
chassis which has been chosen as an example is first taken up, and then 
an explanation is given of the difference in methods which are necessary 
with a different type of construction. 

Moments in a Vertical Plane Perpendicular to the Plane of the 
Struts — Moments in this plane are caused by eccentricities due to the 
fact that the intersection of the struts and spreader tubes, or of the struts 
and wires, does not fall on the line of action of the rubber cord in the 
shock absorber, or to the rise of the axle when side load is present, which 
causes the collar on the axle to bear against the vertical guide plate. The 
manner in which the total moment carried by the front and rear struts 
is apportioned between them is computed by the method of least work. 
As the struts are hinged at the fuselage, with pins running parallel to the 
long axis of the struts, the moment at this point is zero. It increases 
uniformly to its full value at the lower end of the strut. 

Let the total moment going into the front and rear struts on one side 
be unit}'. 

m f = proportion of total moment in front strut. 

m r = (1.00 — m f ) === proportion of total moment in rear strut. 

m x = moment at any point in the struts. 

r f = reaction at upper end of front strut. 

r r = reaction at upper end of rear strut. 

L f = length of front strut. 

L r = length of rear strut. 

If = moment of inertia of front strut. 

I,. = moment of inertia of rear strut. 

153 



Air 


3lane 


Chassis 










Art. 117 








Case 1 














m f 


m f • x 


r r 


m r 




m r • x 






u 


L r 


•> m x 


L r 



Wi 



L, 



m f 2 x 2 dx 
2L f 2 EI f 



w, 



L r 



m r 2 x 2 dx 
2 L 2 EL 



W= w f -(- w r = 



m f < 



L f 2 EI f 



L f m r 2 

x 2 dx -| 

2 L r 2 EI r 



3.2L f 2 EI f 



U 





m, 



3.2L r 2 EI, 



m f 2 L f 3 (1.00— m f ) 2 L r 3 

+ 



6Lf 2 EL 



6 L r 2 EI, 



bv differentiating 



dW 2m f L f 2 (1.00— m f ) L r 







d m £ 6 EI, 



6 EL 



L r / 

m f = / 

EI r / 



U U 

Let = K-, and — 



K. 



EL 



EL 



U L r 1 

+ 

EI f EI r 

= L r I f /(L f I r +L r I f ) 

-K a /(K 1 + K a ) 

Case 2 

In this case the upper ends of the struts are considered to be fixed 
either by a rigid socket or by a pin perpendicular to the long axis of the 
btruts. The moment throughout the strut is constant. 



W = w f + Wi 



L f 



rL, 



m f 2 dx 



2 EL 



m r 2 dx 



2EI r 



r L f 



2 EI f 



+ 



2EI r 

154 



Art 117 



Distribution of Moments 



m f L f (1.00— m f ) 2 L r 

2 EI f ^ 2 EI r 

dW 2m f L f 2(1.00— m f ) L r 



dm f 2 EI f 

L r / 

m f = / 

EI r / 



2EI r 



EI f EI 



K. 



(K, + K 2 ) 



These two solutions show that the distribution of the total moment 
between the struts is independent of the condition of the upper ends of 
the struts. It is also clear that by following the same mathematical 
reasoning the same results would be reached with regard to the distribu- 
tion of a moment in the plane of the struts. It may, therefore, be stated 
that for a moment, either in the plane of the struts or perpendicular to 
it, the distribution between the struts is inversely proportional to their 
length and directly proportional to their moment of inertia. 

Owing to the rigid connection of the struts and spreader tubes to the 
guide plate, any moment in a vertical plane perpendicular to the plane 
of the struts cannot be concentrated in the struts on either side, but is 
transmitted from one set to the other through the spreader tubes. Here 
again the principle of least work will be resorted to. Assume an ex- 
ternal moment of unity applied at B, Fig. 97. 




/ 00 
Externa/ Moment 



Fig. 97 



155 



Airplane Chassis 



Art. 117 



Sum of maximum moments in struts on right side 
m s = m f -\- m r 
m c = Moment in spreader tubes 

= 1.00 — m s = m £l + m n 

— Sum of maximum moments in the struts on the left side. 

As the front and rear struts are the same size, the moment taken by 
each is in inverse proportion to its length. 

m f = = .558m s 



Maximum 
Values 




m r = = . 442 

86.3 

m fl = .558 (1.00— m s ) 

m ri = .442 (1.00— m s ) 

.558 m, 



.01464 m, 




The upper end of each strut, and the left ends of the spreader tubes 
are taken as the origins. The moment in the struts at a distance "x" 
from the origin equals the reaction at that point times "x". 

W = w f + w r -J- w s + w fl + w ri = total work in system. 



r38.l 



w = 



f48.1 



(.01464 m s ) 2 x 2 dx 



2 EI, 



+ 



54.0 



(1 — m s ) 2 dx 
2 EL 



38.1 



(.00916 m s ) 2 x 2 dx 
2 EL 



.01464 2 (1— m s ) 2 x 2 dx 



+ 



2 EI tl 



+ 



156 



Art. 117 



Distribution of Moments 



i.l 



.00916 2 (1— m s ) 2 x 2 dx 
2EI ri 

38.1 



W = 



(.01464 m s ) 2 x^ 
6 EI f 



+ 







(.00916 m s ) 2 x ; ^ 
6EI r 



48.1 



+ 



(1— m s ) 2 x 
2ET S 



54.0 



(0.1464) 2 (1— m s ) 2 x : 



+ 



6EI fl 



38.1 



+ 



dW 



d m. 



(.00916) 2 (1— m s ) 2 x 3 
6EI ri 

2 X.01464 2 m s 38.1' 1 



48.1 



.00916 2 m s 48.1 3 



2 EX- 0375 6 EX- 0375 

2 (1— m s ) 54.0 2 X .01464 2 (1— rn s ) 38.1 s 



2E X- 0210X2 6E X -0375 

2X-00916 2 (1— m s )48.1 3 

=0 

6 EX- 0375 

Since the members are all of the same material E can be cancelled 
from each expression. 
dW 
— = +105.4 m s +83.0m 8 —1285+1285m s — 105.4+105.4 m s 

dm s 

m s = .885 m c =.115 

m f = .494 m fl = .064 

m r = .391 m ri = .051 

In a chassis where, in addition to rigid connections at the guide plate, 
the struts are fixed at their upper ends also, the general procedure is 
the same as in the example just given except that the moment in the 

157 



Airplane Chassis 



Art. 118 



struts is constant throughout their length. The difference in the form 
of the integral is shown in Case 1 and Case 2 in which the distribution 
of the moment between struts is determined. 

If the spreader tubes are hinged, the entire moment is taken by the 
struts on one side. If in this case the struts are fixed at their upper 
ends, the moment in them is constant. 

The processes of integration which have been used in the last few 
examples are as simple as the differentiations previously explained. 

It should be clearly brought out that there are two distinct operations 
performed in the examples given. First, an integration is made to deter- 
mine the value of the work done in each member and in the system, and 
second, the work thus found is differentiated with respect to the un- 
known moment. During the integration the unknown moment is treated 
as a constant while the "x" factor is integrated. Once the integration 
has been made and the limits substituted, "x" drops out and in the dif- 
ferentiation the unknown moment becomes the variable. The example 
which follows illustrates the simple, mechanical operation of integration 
and substitution of limits for the cases that ordinarily arise. 



c x n dx 



n+1 



c (a I1+1 — b n+1 ) 
n+1 



Moments in the Plane of the Struts 

118. Resolution of Moments — Moments in this plane are caused by 
fixed eccentricities of the struts or wires with respect to the center-line of 
the axle slot, or by the variable eccentricity of backward or forward 
thrusts due to the rise of the axle. The distribution of moments in this 
plane between the struts has already been explained. However, as the 
external moments are usually not in the plane of the struts, but in a 
vertical plane parallel to the propeller axis, an explanation must be 
given of the method used in resolving these moments into the proper 
plane. Moments, like forces, can be represented by vectors in magnitude 
and direction, and can be resolved or composed graphically the same as 
forces. In this work the moments are laid off on moment axes so that 
the vectors representing the moments are always perpendicular to the 
plane in which the moments act. Reference to Fig. 98 illustrates the 
problem. Line ab is the vector of the external moment of unity acting in 
a vertical plane. Line ac is the vector of the moment in the plane of the 
struts, while be represents the moment going into the spreader tubes. 
This latter moment causes compression in one tube and a tension in the 
other equal to the moment divided by the spacing of the tubes. 

It has been assumed that the spreader tubes do not assist in taking 
moments that are in the plane of the struts. Some moment is without 
doubt carried by the spreader tubes in torsion, but it is so small that 
it may be neglected. 

158 



Art. 119 



Actual Stresses and Moments in Members 



I. 

II. 

III. 

IV. 
V. 

VI. 



VII. 




Fig. 98. Resolution of Moments 



119. Actual Stresses and Moments in Members. 



Vertical load on one wheel in level landing 
Backward thrust on one wheel in level landing- 
Forward thrust on one wheel in level landing 
Side thrust on one wheel in level landing 
Total moment due to strut eccentricity in side 

view, level landing = .36 X 825 
Total moment due to horizontal eccentricitv 

of wire in side view = 33.5/59.2 X -847 X 

.84 X 1650 
Total moment due to vertical eccentricitv of 

wire in side view = 23.3/59.2 X -847 X -27 



825 
413 
165 
206 



lb. 
lb. 

lb. 

lb. 



-297 in. lb. 



+665 in. lb. 



1650 



= —148 in. lb. 



For a one-wheel landing the loads in I. II, III, IV and V must be 
doubled. The moments caused by the wire eccentricity are calculated 
by finding the horizontal and vertical components of the stress in the 
wire due to a load of unit}' and multiplying by the corresponding eccen- 
tricities. For a one-wheel landing the moments given by V, VI and VII 
ma)- be combined, giving a total moment of — 77 in. lbs. This moment 
is in a vertical plane and must be resolved into the plane of the struts. 
The eccentric moments produced by the thrusts acting through the rise 
of the axle are computed by multiplying the thrust in question by the 
rise in the axle. The latter factor may be determined closely, as fol- 
lows: Assume that the shock absorber will allow the axle to reach the 
top of its slot when a certain dynamic factor, such as 5.0 or 6.0 is ap- 
plied. This factor should be somewhat less than is calculated for new 
cord to allow for deterioration. From this relation can be calculated the 
axle rise for each dvnamic load. Then bv the method of trial the 



159 



Airplane Chassis Art. 119 

dynamic factor which the struts will hold is found, and the axle rise 
computed for this factor. 

In Table XIX is given a summary of the stresses and loads on the 
different members of the chassis for the four important conditions of 
loading. 

Detailed calculations will be made for cases, first in which there are 
moments only in the plane of the struts, and then in which moments are 
present both in the plane of the struts and perpendicular to this plane. 
Rear Strut, Case 4 (one wheel landing with .5 back thrust). 
Calculation of moment due to rise of axle. 

Total rise of axle for a load factor of 6.0 in level landing=5.12 in. 

Rise of axle for a load factor of 1.0 in one wheel landing = 1.71 in. 

Back thrust=825 lbs. for a load factor of 1.0 in one wheel landing. 

Assume the factor of safety to be 2.1. The maximum eccentricity of 

back thrust then equals 2.1 X 1.71 = 3.59 in. Total eccentric moment 

due to axle rise = 825 X 3.59 = +2960 in. lbs. The portion of this 

moment going into the rear strut = 2960 X -442 = +1310 in lbs. (See 

p. 156.) 

To resolve this moment, which is in a vertical plane, into a moment 
in the plane of the struts, it must be multiplied by the factor 1.10 (See 
Fig. 98) + 1310 X 1.10 =+1440 in. lbs. 

This moment must then be combined with that due to the strut and 
wire eccentricities, so that the final moment = +1440 — 37 = +1403 
in. lbs. 

1403 

f b = =15,400 

.0912 

(1010 + 695) 

f c = = 7,490 

.228 

f t = 22,870 lbs. per sq. in. 

Calculation of allowable unit stress. 

P/A = 36,000 - 0.37 (85) 2 = 33,300 lbs. per sq. in. in straight com- 
pression. 

Ultimate allowable stress in bending = 55,000 lbs. per sq. in. 

The formula used in obtaining the ultimate allowable stress in com- 
bined bending and compression is the same as that given in Art. 174 for 
the stresses in spars. 

F a = (F-C) +C 

f b + f c 
15,400 

F a = (55,000 - 33,300) + 33,300 = 47,900 lbs. 

15,400 + 7470 persq.in. 

47,900 

F.S. = = 2.1 

22,870 

160 



Art. 119» 



Actual Stresses and Moments in Members 



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Airplane Chassis Art. 119 

This factor of 2.1 applies to the vertical component of the load, for 
convenience taken as the weight of the airplane. (To obtain the true 
factor of safety based on the assumption that the resultant load, ob- 
tained by combining the vertical and horizontal loads, equals the weight 
of the airplane, the factor 2.1 must be multiplied by 1.115 which equals 
VT50^TT700 2 , F.S. = 2.1x1.115 = 2.33). A similar factor must be 
applied to the other factors of safety which are given in Table XIX. 

At their lower ends the struts are rigidly connected to a vertical steel 
plate. At the upper ends they are connected to the longerons by pins 
in the plane of the long axis of the strut. In computing the effect of 
column action caused by the direct stress, a fixity factor of 3 was used 
in the parabolic column formula of Art. 24 when the bending moments 
on the struts were in the plane of the struts. This value of the fixity 
coefficient, rather than the value of 4, was adopted for the reason that 
although the struts were assumed to be fixed in their own plane, it is 
practically impossible to obtain theoretical fixity. When the main bend- 
ing moments were perpendicular to the plane of the struts, the fixity 
factor was reduced to 2 on account of the direction of the hinge pins at 
their upper ends. 

The solution for a case in which there are external moments about 
both axes will now be considered. The side thrust, acting through an 
eccentricity equal to the rise of the axle, produces a moment which is 
treated in the same manner as that caused by the back thrust in Case 4. 

Rear Strut, Case 7 (Level landing with .25 side thrust) 

Calculation of moments perpendicular to plane of the struts caused 
by rise of the axle. 

Rise of axle for load factor of 1.0 = .853 in. 
Side thrust = 413 lbs. 
Assume that the factor of safety is 4.3. The maximum eccentricity 
of the side thrust then equals 4.3 X -853 = 3.67 in. 

Total eccentric moment due to axle rise = 413X3.67=1515 

in. lbs. 
The portion of this moment going into the rear strut = 1515 X 
.391 = 592 in. lbs. (See p. 157.) 
Location of most stressed fiber. 

m 1 = — 145 in. lbs. = moment in the plane of the struts. 
m 2 = 592 in. lbs. 

mi/m = .245 from Fig. 82 y = .495 X 1.125 = .556 in. 

x = .l50X 1.125 = .169 in. 
145 X -169 

f bl = = 280 

.0872 

592 X .556 

f b „ = = 8750 

.0375 

162 



Art. 119 Actual Stresses and Moments in Members 



(499 + 9) 

f c = = 2230 

.228 

f t = 11,260 lbs. per sq. in. 

Calculation of allowable ultimate stress. 

P/A = 36,000 — .56 (128) 2 = 26,800 lbs. per sq. in. in straight. 

compression. 
Ultimate stress in bending = 55,000 lbs. per sq. in. 

9,030 

f a = (55,000-26,800) +26,800 = 49,400 lbs. per sq. in. 

11,260 

49,400 
F.S. — = 4.4 

11,260 

Calculation of Spreader Tubes. 
Case 7. 

Length = 54.0 in.; L/p = 163 
Assume pinned ends 

9.86X30,000,000 

P/A = — 11,100 lbs. per sq. in. 

163 2 

Compression in the two tubes = 413 lbs. 
Assume that the axle rises to the top of the slot. 
Portion of moment due to axle rise going into spreader tubes = 
413 X 5.12 X .115 = 243 in. lbs. (Seep. 157.) 



243 

f„ = = 2890 

.084 

413 
f c = = 1080 

.384 

f, = 3970 

2890 

f ;l = (55,000— 11,100) + 11,100 = 43,100 

3970 

43.100 

F.S. = = 10.8 

3970 

163 



Airplane Chassis 



Art. 119 






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164 



Art. 120 Hinged Axle Chassis 



in. O.D., elliptical tubing, .049 in. gage and 1.7 
fineness ratio. Their properties are as follows: 

Area = .228 

Moment of Inertia= .0375 about long axis 

Moment of Inertia= .0872 about short axis 

Section modulus = .0666 about long axis 

Section modulus = .0912 about short axis 

Radius of gyration = .406 about long axis 

Front Wire 1 — 5700 lb. streamline wire; area = .0376 

Spreader tubes: 2 — 1 in. O.D. .065 in. gage round tubes 

Total area = .384; total I = .042; P = .331; total I/y = .084 

m t is moment in plane of struts. 

m 2 is moment perpendicular to plane of struts. 

L/p for front strut is 103 about long axis 
67 about short axis 

L/p for rear strut is 128 about long axis 
85 about short axis 

Parabolic column formula for 55,000 lbs. per sq. in. steel is 
P/A = 36,000 - .56 (h/ P y- for C = 2 
P/A — 36,000 - .37 (L/p) 2 for C — 3 

In computing this table it is assumed that the shock ab- 
sorber has a service factor of 6.0. 

It will be observed from a study of Table XIX that the greatest 
stresses are in general produced by bending moments and not by direct 
compression. It will also be noted that the moments caused by back 
thrusts acting through the rise of the axle are of opposite sign to the 
moments resulting from the eccentricity of the struts. The effect of this 
is especially important in Case 1. Therefore, it is suggested that an 
eccentricity of the struts in side view may be advantageous in neutral- 
izing the moments produced by back thrusts. In Case 6, however, the 
forward thrust produces a moment which is in the same direction as that 
caused by the strut eccentricity. It is evident that in determining the 
eccentricity to give minimum stresses, both Cases 4 and 6 must be con- 
sidered. 

It is further suggested that the front strut, which usually is much 
more severely stressed than the rear strut, be made of a heavier section 
than the rear so that the factors for both struts will be nearly equal. 

120. Hinged Axle Chassis — As a number of chassis are designed 
with the axle hinged at two points instead of being continuous, some of 
the features of such a design will be discussed. In what is probably the 
best type, the elastic cord is wrapped over the axle on each side of the 
struts and under the struts; the spreader tubes are secured to the struts 
by a pinned connection; the axle is hinged to a cross member between 
the spreader tubes, and at this point the latter are supported by a wire 
taken off the small cross tube. Fig 99 shows these features diagram- 
matically. With this design no bending moment is put into the spreader 
tubes, nor any moments in the struts except those due to strut eccen- 

165 



Airplane Chassis 



Art. 120 



tricities. The load on the struts is increased by the amount of the stress 
in the wire which takes the reaction from the axle at its hinged end. 
However, this affects only slightly the design of the struts, and the 
chassis as a whole will be much lighter for the same strength than the 
other types of hinged axle chassis. Referring to Fig. 99 the vertical 
component of the load transmitted by the elastic cord to the struts equals 
W •a/b, and the reaction at c that is carried by the support wire equals 

(a-b) 

W . The fact that the load which the shock absorber has to carry 

b 
is increased by the amount of this reaction is one of the most disadvan- 
tageous features of all hinged axle designs. 

If the wire at c is omitted, the spreader tubes must carry in bending 
the axle reaction at that point. This adds largely to their weight. To 
have them rigidly connected to the struts would partly relieve the bend- 
ing moment on the tubes. But, as the moment would go directly into 
the struts and have to be carried by them in bending about their long- 
axes, such a rigid connection would not be at all advisable. 




Fig. 99. Chassis with Hinged Axles 

166 



Art. 121 



Problems in Chassis Design 




-Loop of 
Elastic Cord 

Fig. 100 

Fig. 100 is a line sketch illustrating a design in which the elastic cord 
is attached to the spreader tubes, which pass through the struts, rather 
than directly to the struts. Since the spreader tubes are subjected to a 
considerable bending moment their diameter and weight have to be in- 
creased. There is also danger that some of the moment from the tubes 
will be transmitted into the struts. The wire at a to take the axle re- 
action may be omitted, though only by largely increasing the size of 
the tubes. One advantage with this type of chassis is the general clean- 
ness of design and the good streamlining that is possible. The distance 
between the centerlines of the wheel and shock absorber can be made 
less than in the more usual type, thereby decreasing the bending moment 
on the axle and, hence, its size. The remarks made in Art 122, on the 
design of axles, relative to the loss in strength caused by heating for the 
purpose of brazing, as to the quality of steel used, and to the piercing 
of a member by holes for a pin, etc., are applicable to spreader tubes in 
hinged axle chassis, which are heavily stressed in bending. In some 
tests on this type of chassis failure has first occurred in the spreader 
tubes because these principles were not adhered to. 

Chassis with hinged axles have been somewhat of a novelty, but de- 
signers are in general more inclined to use the one-piece axle. The dif- 
ference between the two types, in so far as their riding qualities are con- 
cerned, is negligible, while the design using the one-piece axle is simpler 
and generally stronger for the same weight. 

121. Problems in Chassis Design — General cleanness of design 
should be the aim in chassis construction. This and other desirable 
characteristics ma}' be obtained by meeting the following requirements: 

I. Adequate but not excessive strength. 

II. Minimum weight. 

III. Low structural air resistance. 

IV. Simplicity of construction. 

V. Easy riding qualities. 

167 



Airplane Chassis Art. 121 

The various problems in chassis design are concerned with and affect 
these factors. 

In connection with the first point, the question of suitable factors of 
safety will be discussed elsewhere. It should be stated here, however, 
that the strength of the struts and wires should be as uniform as possible. 
In general, the axle, and, in the case of hinged axles, the spreader tubes, 
should be no stronger than the weaker of the two struts. It is per- 
missible to have the factor on the shock absorber no higher than on the 
axle. As the whole object of a chassis is to break the fall of the main 
portion of the airplane, there should be a progressive failure of the va- 
rious component parts: first, the shock absorber, then the axle, and 
lastly the struts. The struts, too, should fail before the load becomes 
heavy enough to crush in the fuselage or longerons. However, but a 
small difference in factor is required to effect this. In many chassis 
tests the shock absorber has failed at a factor of 3.0, the axle at 4.5, and 
the struts at 8.0. Such a wide variation in strength indicates poor de- 
sign. 

Minimum weight is secured, partly by seeing that no members have 
excessive strength, and partly by the elimination or reduction of eccen- 
tricities that cause large stresses. It may be noted again that eccen- 
tricities producing moments in a plane perpendicular to the plane of the 
struts are much more serious than eccentricities producing moments in 
the plane of the struts. The following method is suggested for reducing 
the eccentricity produced by the rise of the axle, which, when a back 
thrust is present, causes a moment in the plane of the struts. As is 
shown in Fig. 101, the axle should be dropped vertically below the inter- 
section of the struts and a line through the centers of the spreader tubes. 
The axle fairing should be lowered till it rests on top of the spreader 
tubes, and the axle till it touches the bottom of the fairing. When the 
axle is forced up in the slot, the eccentricity of the back thrust is re- 
duced by the amount the axle is dropped. This feature will also lessen 
the height of the unsupported part of the vertical guide plate, and, hence, 
make it more rigid in resisting side thrust loads when the axle is forced 
nearly to the top of the slot. 

The careful streamlining of shock absorbers by aluminum fairing, 
and of the axle either by this or by suitable wood fairing, the use of the 
standard strut section described in Art. 91 for the struts whether they 
are all wood or are faired steel tubing, the elimination of wires in the 
plane of the rear struts where practicable and the streamlining of those 
used, care in cleaning up fittings and minor details, and the attachment 
of the wheel fairing disks to the tire at its maximum diameter instead of 
to the rim of the wheel should largely reduce the structural resistance 
of a chassis. 

In designing, an effort should always be made to eliminate compli- 
cated fittings and those which present difficulties in manufacture. As 
a rule, simplicity and strength in construction go together to give de- 
pendability and long service. The article dealing with shock absorbers 

168 



Art. 122 



Design of the Axle 




Spreader Tubes 



Fig. 101 




6 £5 lbs. 



Fig. 102 



discusses the resiliency of elastic cord and its effect on the easy riding 
qualities of a chassis. 

122. Design of the Axle — Axles are calculated for the vertical load 
in level landing, and are designed for a factor equal to or slightly less 
than the factor for the worse stressed struts under the same condition. 
In a one-wheel landing the factor is cut in half, and with a back thrust it 
is still further reduced, perhaps even more than the factor for the struts. 
However, if an axle is overstressed it merely bends and no damage is 
caused to the airplane as a whole. Furthermore, such failure can be 
readily detected and the axle easily replaced. Under no circumstances 
should any part of an axle be heated as high as a low red heat after it 
has once been given its final heat-treatment, or its strength will be de- 
creased to about 1/3 its proper value. Also, soft steel or plain high 
carbon steel should never be used in axle design. Alloy steel of an ulti- 
mate tensile strength of 200,000 lbs. per sq. in., conforming to U.S.A. 
Signal Corps Specification No. 10,229, can be obtained and depended 
upon. According to this specification, all axles, to pass inspection, must 
be subjected to a bending stress of 180,000 lbs. per sq. in. without injury. 

Fig. 102 gives the necessary data for the design. 

The bending moment on the axle in level landing 



O.D. of axle 
F.S. 



= 825 lb. X 4.94 in. = 4075 in. lb. 
\y 2 in.; Wall = 3/32 in.; I/y = .137 
4075 

= 29,700 lb. per sq. in. 

.137 
200,000 



6.7 



29,700 
123. Design of the Shock Absorber — There are two general types 
of shock absorbers; one in which the cord is wound directly over the 
axle and under the struts or spreader tubes, and one in which the cord 
is wound on spindles, the lower spindle being attached to the struts 
or spreader tubes and the upper one to the axle. The former type is no 

169 



Airplane Chassis 



Art. 123 



longer used in careful, refined design for the following reasons: 
I — Owing to the way in which the cord is wound it is impossible to de- 
termine with accuracy its elongation for any rise of the axle, and there- 
fore to calculate the depth of slot required. II — As the cord is 
wrapped in several layers the chafing and cutting of the under layers by 
the upper when the cord is under strain causes rapid deterioration. 
Ill — For the same reason, the distribution of load between the cords is 
unequal.' Some cords are likely to take enough load to break the fabric. 
IV — With this type of shock absorber streamlining of the cord is very 
difficult. The later designs of spindle shock absorbers permit the cord 
to be wound on the spindles when they are removed from the chassis. 
As the cord is wrapped in a single layer chafing and unequal load dis- 
tribution are a minimum. The elongation of the cord is easily com- 
puted. Because of the compactness and small section of a spindle 
shock absorber, it is readily streamlined and offers but little resistance. 
Fig. 103 shows the absorber used on the chassis which is taken as the 
example in this analysis. 

The function of a shock absorber is to lessen the shock of landing by 
absorbing the kinetic energy of the airplane which is falling with a cer- 
tain vertical velocity. The energy thus taken up by each strand of cord 
equals the elongation of the cord times the average tension in the cord 
during this elongation. For example, assume that the airplane used in 
this chapter, whose weight is 1650 lbs., falls freely through a distance 
of 10 in. The kinetic energy of the airplane at the instant of contact 
= W = 16,500 in. lbs. It is assumed that 1/2 in. cord whose average 




Fig. 103 

170 



Art. 123 



Design of the Shock Absorber 



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Fig. 104. Typical Load Elongation Curve for J/2-in. Rubber Cord 




Airplane Design Art. 123 

load-elongation curve is shown in Fig. 104 is used on the airplane. 
Below are given the necessary data for the shock absorber. 

One Unit 

Length of 4 partial loops = 14.1 in. 

Length of 1 complete loop = 2(1.67+3.21) = 9.76 in. (See Fig. 105) 

Number of complete loops = 13 

Total length of unstretched cord 

= 13x9.76+14 =141.1 in. 

Total possible rise of axle = 5^/g in. 

Total elongation of cord = 5.12 X 30 =153.6 in. 

153.6 

Percentage maximum elongation = =109 per cent 

141.1 

In the two shock absorbers on the airplane there are 2 X 30 = 60 

16,500 

strands of cord. The enersw to be taken up bv each strand = = 

60 

275 in. lbs. The distance through which the tension in each strand acts 

153.6 

per 1 per cent elongation = = .0470 in. Referring to curve 

30 X 109 
B , Fig. 104, which is obtained by replotting curve A, using for ordinates 
actual distances instead of percentage elongations, the area under the 
curve to the left of any ordinate, when multiplied by the scales of force 
and distance, equals the work done on the cord in stretching the given 
distance, or the amount of the energy absorbed. When the stretch 
equals 3.65 inches, corresponding to an elongation of 76 per cent, the 
area under the curve equals 6.87 square inches. 

Energy = 6.87 X 40 X 1 = 275 in. lbs. The totatl energy which can 
be absorbed by each strand before the axle rises to the top of its slot at 
an elongation of 109 per cent equals 557 in. lbs. Therefore, the max- 
imum capacity of both shock absorbers = 557 X 60 = 33,500 in. lbs. 

This study of the resiliency of elastic cord shows the importance of 
the character of the load-elongation curve. Cords which may require 
the same load to produce a 100 per cent elongation will give very dif- 
ferent results in service. The cord of which C is the load-elongation 
curve, is so weak and has such a small resiliency that it is practically 
useless. On the other hand, the cord represented by curve D is too 
stiff to give easy riding qualities. The stress in the cord for a given 
resiliency is much greater than with the more normal cord represented 
by Curve A. The areas indicated in Fig. 104 under the two curves are 
the same, 275 in. lbs., but the tension in the cord at e is more than 20 
per cent greater than in the cord at /. It is clear, therefore, that to 
absorb a given amount of energy with cord D will put a greater stress 
in the struts with the average landing. On the contrary, when the land- 

172 



Art. 123 Design of the Shock Absorber 



ing is so severe as to cause an elongation greater than about 95 per cent 
the struts will be less stressed with cord D. 

The usual basis for determining the number of strands of cord and 
the necessary rise of the axle, where the cord to be used conforms to a 
standard specification, is that the total load required to force the axle to 
the top of the slot (against the resistance of the cord) divided by the 
weight of the airplane shall equal, or nearly equal, the factor of safety 
of the worse stressed struts in a level landing. 

From curve A the load required to elongate the cord 109 per cent is 
186 lbs. 

186X60 

F.S. = = 6.75 

1650 

Owing to the latitude in regard to the stiffness of cord that is neces- 
sary in the specification for elastic cord, cord which would meet the 
lower limit set by the specification would give a factor of safety of 6.25, 
which is satisfactory. 

The requirements of good design limit the range of maximum elonga- 
tion. This should be between 100 and 120 per cent. A value of 110 
per cent will ordinarily give the best results, for if the maximum elonga- 
tion is much less the design is not economical, since, owing to the steep- 
ness of the load-elongation curve for elongations greater than 100 per 
cent, a slight increase in elongation gives a large increase in load carried, 
while if the elongation is much more the action of the cord is too stiff, 
and the cord is liable to deteriorate rapidly. There are two ways to se- 
cure any desired elongation: one is to use a large amount of rubber and 
have the axle slot very deep; the other is to decrease to a minimum the 
distance between the spindles of the shock absorber. The latter method 
gives a small, compact design, but, because of the small amount of 
rubber used, it lacks resiliency or shock absorbing power and is not, 
so effective in reducing the shocks of bad landings. 

A moot point in connection with shock absorbers is the advisability 
or necessity of putting a large initial tension in the cord. Whether it is 
necessary to employ such a tension depends both on the quality of the 
cord and the design of the shock absorber. If a cord is weak, especially 
at low elongations, and has the general characteristics shown by curve 
C, Fig. 104, it will stretch so much, unless given a large initial tension, 
that the axle will rise an inch or so even under the dead load of the air- 
plane. Or if the design does not provide a sufficient number of strands 
of cord to support the dead load without causing a large tension in each 
strand and consequently a large elongation, the axle will rise an exces- 
sive amount unless an initial tension is used. One important fact should 
be brought out regarding the effect of initial tension in reducing the 
resiliency of the shock absorber. If, for instance, the cord represented 
by curve D is wound with a tension of 50 pounds it will be initially 
elongated 50 per cent, and all the capacity for absorbing energy repre- 
sented by the area under curve D to the left of the ordinate at 50 per 

173 



Airplane Chassis Art. 1-23 

cent elongation will be lost. A similar loss will, of course, occur when 
initial tension must be used because of an insufficient number of strands, 
even though the cord be of good quality. Another point to be men- 
tioned is that unless a spring balance is used when the cord is being 
wound, or unless the elongation of the cord for the desired tension is 
known and the cord cut to the proper length before winding, there is no 
way of knowing what initial tension is actually used. The cord may be 
stressed twice as much as was intended. In spite of these facts it may 
be advisable to use an initial tension of 30 to 50 pounds where the de- 
sign, owing to certain limitations, could not provide an adequate num- 
ber of strands of cord. A distinction should be made between an insuffi- 
cient number of strands of cord and an insufficient amount of cord. In 
the first case the amount of cord may be ample, but inefficiently dis- 
tributed, the spindles probably being too far apart. 

In general, a shock absorber should be originally designed, first for 
a good quality cord that meets the government specifications, and sec- 
ond for a small initial tension of only 10 — 15 lbs. or just enough to wrap 
the cord snugly. This will lengthen the life of the cord because, not 
being continually under a high stress, it will not chafe badly. If, also, 
enough cord is used so that the cord is very seldom stretched nearly up 
to the breaking of the fabric it should give good service. 

Government Specifications. 

Load at 100 per cent elongation 

3/8 in. cord 65— 90 lbs. 

1/2 in. cord 145—180 lbs. 

5/8 in. cord 240—275 lbs. 

124. Strength Factors for Chassis — The factors given in column 3 
are applicable only to the struts. They represent the requirements of 
the present standard static test for chassis. In this test the load is 
equally divided between the wheels, and the testing jig is so arranged 
that the resultant load makes an angle with the normal to the propeller 
axis equal to the angle corresponding to a on page 176 for the airplane 
in question. For most designs this is equivalent to considering the 
propeller axis horizontal, and combining a vertical load with a back 
thrust, equal to from .2 to .25 of the vertical load, so that their resultant 
equals the weight of the airplane. 

The minimum factor for loadings (4) and (6), given in Art. Ill, 
shall be not less than 35 per cent of the required factor for the static 
test, which is specified in Table XX. For the fully loaded day and night 
bombers the minimum factor for loadings (4) and (6) shall be not less 
than 40 per cent of the factor in Table XX. 

When the above strength factors for the struts are computed an- 
alytically, the stresses upon which such factors are based must include 
stresses due to the various eccentric moments, which have been taken 
up in detail in the present chapter. The methods used to calculate the 

174 



Art. 124 



Factors of Safety 






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75 



Airplane Chassis Art. 125 

stresses should closely follow those which have been previously dis- 
cussed in this chapter. 

The axle will be required to sustain, without permanent deformation, 
a load corresponding to the factor specified in column 4. What may 
be considered failure of the axle shall not occur at a factor less than the 
factor specified in column 5. In the case of hinged axle chassis these 
same factors shall apply to the spreader tubes where they carry major 
stresses. 

The axle shall not be forced to the top of its slot by a load less than 
that corresponding to the factor given in column 6. Furthermore, the 
outside fabric of the elastic cord shall not be injured by this load. 

125. Relation of Chassis to Fuselage — The height of a chassis is de- 
termined, first, by the clearance of the propeller with respect to the 
ground when the propeller axis is horizontal, and second, by the angles, 
<£ and 0. In measuring these angles the propeller axis is considered 
horizontal. <f> is the angle formed by the ground-line and a line through 
the tip of the tail skid and tangent to the lower part of the wheel; is 
the angle between the ground-line and a line through a bottom wing tip 
and the lower part of the wheel on the same side as the wing tip; this 
angle shows in front elevation. The minimum value of the propeller 
clearance may be taken as 10 in. The longitudinal location of the 
chassis is fixed by the angle a between the vertical through the point of 
tangency of the wheels and the ground, and a line through this point 
and the center of gravity of the airplane. The usual range of values 
of <£, and a for a number of types of American and foreign airplanes 
is given below. 

TABLE XXI 

Single Seater Two Seater Heavy Bombers 

<f> 13°— 14° 11°— 12° 10°— 11° 

8 19°— 22° 13°— 17° 9°— 12° 

a 15°— 18° 10°— 13° 9°— 12° 

T/S 17 —.21 .12 —.15 .16 —.20 

In each class, the value of a depends on the minimum speed of the 
airplane. It is greater for high landing speeds. Values of T/S or the 
ratio of the tread to wing span are useful in determining the tread. The 
slopes of the chassis struts are determined largely by the location of 
suitable bulkheads, to which the upper ends of the struts may be at- 
tached. If possible, the front struts should support the engine directly. 



176 



CHAPTER VI 

CONTROL SYSTEMS 

126. Introduction — The application of the principles of structural 
analysis to the design of control systems is not so general as in the case 
of wing structures. As a consequence, failure of controls during a sand 
test on control surfaces has frequently occurred, sometimes at a very 
low factor. So designers, knowing of such failures, have often added 
unnecessary weight to the controls to insure adequate strength. 

Both control surfaces and systems may be designed to withstand 
air loads which are determined either by what is considered the max- 
imum force it is possible for a pilot to exert, or by definite unit loads 
fixed by specification for each type of machine. The basis for specifying 
these latter loads will be discussed in Chapter VII. It is evident that 
designs based on such loads will be more uniform and better balanced. 

In the present chapter, an investigation will be made of the control 
system for a single-seater pursuit airplane. The functioning of this sys- 
tem is shown diagrammatically in Fig. 106. The design, though a 
reasonable one, might be considerably improved. It is impossible, of 
course, to proportion each tube and bolt so that all parts are stressed 
up to the limit. However, the weakest part of a control system should 
be somewhat stronger than the minimum factor required for the control 
surfaces. A margin of about 10 to 15 per cent should be allowed. Sev- 
eral special cases will be considered to illustrate certain features not 
occurring in the controls of this pursuit airplane. 



STRESS ANALYSIS FOR CONTROLS OF PURSUIT AIRPLANE 

127. Calculation of Moments — A torque system of control, shown 
in Fig. 106, is used for the ailerons; and for the elevators, instead of the 
usual two masts, there is only one attached to the elevator tubes inside 
the fuselage. The curved rudder bar, Fig. 114, is a trifle unusual, but 
aside from these features the controls conform to common practice. 

The first step in the design is the calculation of the moments pro- 
duced by the loads on the elevators, ailerons and rudder. For a single- 
seater pursuit the present Government specifications require an ulti- 
mate load on the horizontal tail surfaces which shall average 35 lbs. 
per sq. ft.; on the vertical tail surfaces an average load of 30 lbs. per 
sq. ft.; and on the ailerons the same loading as on the elevators. The 
distribution of load between the fixed and movable surfaces, both 
horizontal and vertical, is shown in Fig. 107. Owing to the rounding 
of the ends of the surfaces, the center of gravity of the load on each 
must be calculated. This is done by dividing the surface into a sufficient 
number of strips. As the loading for the movable surfaces is triangular, 
the center of gravity of the load on each strip will be 1/3 the length 
of the strip from the leading edge of the surface. 

177 



Control Systems 



Art. 127 







178 



Art. 127 



Calculation of Moments 




Fig. 107. Load Distribution Between the Fixed and Movable Surfaces 

Calculation of Unit Loads. 

I. Elevator. 

Area of elevators = 16.7 sq. ft. 

Area of stabilizer = 12.5 sq. ft. 

Let "w" equal unit load on stabilizer. 

4 w 
12.5 wH X— X16.7= (12.5 + 16.7) 35 = 1020 lbs. 

3 2 

w = 43.3 lbs. per sq. ft. 
2/3 w = 29.0 lbs. per sq. ft. 

= Average elevator and aileron load. 

II. Rudder. 

Area of rudder = 8.6 sq. ft. 

Area of fin = 3.8 sq. ft. 

Let zv equal unit load on fin. 

3.8 w + 4/3 X .5 w X 8.6 — (8.6+3.8) 30 — 372 lbs. 

w = 39.0 lbs. per sq. ft. 
2/3 w = 26.0 lbs. per sq. ft. 
= Average rudder load. 

Calculation of Center of Gravity of Elevator Load. 

(23.5X30.2) 7.83 + (3.75x23.15) 7.72+(3.55x21.8) 
7.27+(3.4Xl8.9) 6.3 + (1.9Xl4.2) 4.72+(4.46X 
23.15) 7.72+(3.7x21.3) 7.10+(42) 6.0 



1200 



179 



Control Systems 



Art. 128 



8993 

x = = 7.5 in. = arm of load 

1200 

Calculation of Moments. 
I. Elevators. 

Total moment = Area X average load X mean arm 
= 2 X 8.33 X 29.0 X 7.5 
= 3620 in. lbs. 



II. Rudder 

Moment 



Area X average load X mean arm 
8.63 X 26.0 X 9.45 
2120 in. lbs. 



III. Ailerons 

Total moment = Area X average load X mean arm 
= 2 X 12.5 X 29.0 X 7.12 
= 5160 in. lbs. 



128. Investigation of Control Stick for Elevator Loads — Refer to 
Fig. 108. 



5 W 



Torque 7l/be 



E/evolor Cob/e 




£/evotor Cob/e 
Fig. 108. Control Stick 



The moment on the control stick from the elevator or aileron 
loads, or on the rudder bar from the rudder loads is not necessarily the 
same as the moment of these loads about the hinges of the movable sur- 
faces, but is dependent on the relation between the height of the control 
masts and the length of the lever arms on the control stick or rudder 
bar. In the design under consideration the height of the control masts 
on the elevator is \y 2 ins. 

180 



Art. 128 Investigation of Control Stick for Elevator Loads 

Moment of elevator load = 3620 in. lbs. 
3620 

Reaction at grip = = 173 lbs. 

21 

Maximum moment in single tube = 173 X 16 = 2770 in. lbs. 
Tube size = 1% in. O.D., 3/32 in. wall. 
Section modulus = .0724 
2770 

.f b = = 38,300 lbs. per sq. in. 

.0724 

55,000 

Coefficient = = 1.43 (satisfactory) 

38,300 
A wall thickness of 1/16 in. gives a maximum bending stress of 52,800 
lbs. per sq. in., which does not provide sufficient margin when 55,000 lbs. 
per sq. in. steel is employed. A higher strength alloy steel may be used 
and the gage reduced, provided the tube is not annealed by brazing so 
as to destroy the effect of the heat treatment. 

3620 

Pull on elevator cable = = 805 lbs. 

4.5 
Elevator cable is 1/8 in. 7 X 19 extra flexible, with 2000 lbs. 
capacitv. 

2000 

Coefficient = =2.48 (satisfactory) 

805 
Shear on 3/16 in. nickel steel pin holding elevator cable. 

Area of 3/16 in. pin = .0276 sq. in. 
805 

f s = =14,600 lbs. per sq. in. (satisfactory) 

2 X .0276 

Bearing on 3/16 in. pin. 

805 

f c = = 17,200 lbs. per sq. in. (satisfactory) 

.1875 X-25 

Moment on control stick at section X — X, Fig. 108, 
— 805 X 4.5 = 3620 in. lbs. 

1.375 3 X -25 .312 X .25 

Xet I of section = = .0535 

12 12 

3620 X -688 

f b = = 46,600 lbs. per so. in. 

.0535 

181 



Control Systems 



Art. 128 



55,000 

Coefficient = = 1.18 (satisfactory) 

46,600 

Shear on 5/16 in. bolt in control stick 
Load = (805 + 173) = 978 lbs. 
Area = 2 X .0766 sq. in. 



978 



2 X .0766 



6380 lbs. per sq. in. (satisfactory) 



Bearing of 5/16 in. bolt 

Bearing area = 2 X -312 X -125 = .078 



978 



= 12,500 lbs. per sq. in. (satisfactory) 






.078 

The short torque tube to which the control stick is attached is held 
in place by a 3/16 in. bolt passing through a collar on the tube. 

Shearing stress on this bolt 

978 



f. 



2X .0276 
Bearing stress on this bolt 
978 



17,700 lbs. per sq. in. (satisfactory) 



t 



2X.187X-0625 



41,700 lbs. per sq. in. (satisfactory) 



-Bearing 3/ocAc 
brozed fo tube 



3razc</ on 




Fig. 109. Section Through Control Stick at Torque Tube 

182 



Art. 129 Investigation of Control Stick for Aileron Loads 

129. Investigation of Control Stick for Aileron Loads — Reference 
should be made to Fig. 109, which shows a section through the control 
stick and torque tube. The total aileron moment is 5160 in. lbs. 

Shear on 5/16 in. bolt in transmitting this moment 
5160 

Shear = = 3750 lbs. 

1.375 
3750 

f s = = 48,900 lbs. per sq. in. (satisfactory) 

.0766 

Bearing of 5/16 in. bolt on control stick 
5160 

Compression = =3300 lbs. 

1.562 
3300 

f c = = 56,500 lbs. per sq. in. (satisfactory) 

.312 X .1875 
This value is somewhat high, but bearing stresses may be greater 
than the ultimate strength of the material without harm. 

Compressive stress in sides of control stick 

Compression = 3440 lbs. 
Compressive stress on net area through 5/16 in. bolt hole= 

3400 3440 

f c = = 

(.125X1.062+. 0625 X- 437) .160 
= 21,500 lbs. per sq. in. (satisfactory) 

As the sides of the control stick are unsupported for a length of 
3y2 ins., there will be some column action and consequently a reduction 
in the allowable ultimate stress. Since a careful determination of the 
correct average value of the radius of gyration of this column would be 
very difficult, the extreme assumption will be made that the column sec- 
tion is a flat plate instead of a varying circular arc. 

3440 

f c — = 20,000 lbs. per sq. in. 

.172 

31,700 

Coefficient = =-1.58 (satisfactory) 

20,000 

Area = .125 X 1.375 = .172 

h .125 

P =^— = =.0361 

V12 3.46 

L/ P =97 

f c = 55,000— 240L/ P = 31,700 lbs. per sq. in. 

183 



Control Svstems 



Art. 130 



130. Investigation of Torque Tube to Which Control Stick is Secured 

Torsional shear stress 

Aileron moment — 5160 in. lbs. 
Polar I of l^X7s2 in- tube = 2X-0407 = .0814 
5160 X -562 

f s = =35,700 lbs. per sq. in. 

.0814 
See Fig. 153 for Ultimate Torsional Shear Stress 
44,000 

Coefficient = = 1.23 (satisfactory) 

35,700 
Bearing of 5/16 in. bolt through joy stick on tube. From Fig. 109 it 
will be observed that a bearing block is brazed to the torque tube. 

Distance of center of bearing area from center of tube = 
(.469+.5X-219)=.578 in. 
5160 
Force = = 4470 lbs. 



2 X .578 
4470 



65,400 lbs. per sq. in. 



.312 X .219 
This value in bearing is high, but failure will not occur. 

131. Investigation of Aileron Lever — Refer to Figs. 110 and 111. 

5160 

Vertical force on outer end of arm = 397 lbs. 

2X6.5 

Since the rods connecting the ends of the arm with the bell cranks 
make an angle of 16 degs. with the vertical, a horizontal component is 
present which produces a moment about the long axis of any cross- 
section of the arm. 

Vertical component = 397 lbs. 

Horizontal component = 397 X tan 16° = 114 lbs. 

Moment about horizontal axis = 114 X 6.5 = 740 in. lbs. 

Moment about vertical axis = 397 X 6.5 = 2580 in. lbs. 

H 

6 




Fig. 110. Aileron Lever on Torque Tube 
184 



Art. 132 



Investigation of Bell Crank Lever 




Fig. 111. Section A-A Through Aileron Lever 



By the method of trial it is found that "B" is the most stressed point 
on the cross-section "A — A", Fig. 111. 

2580 X -90 
f bl = = 31,300 



f b2 



.0741 
740 X -14 

.0281 



3,700 



Resultant stress = 35,000 lbs. per sq. in. 

. 55,000 

Coefficient = = 1.57 (satisfactory) 

35,000 

Stress on weld between arm and torque tube 

Minimum area of weld = ttX 1.125 X2X-049 = .346 sq. in. 

5160 
Total stress on weld = = 9160 lbs. 



9160 



f a = 



346 



.5625 
26,500 lbs. per sq. in. 



This is a dangerously high stress on a welded section due to the un- 
certainty of the quality of the weld. 

132. Investigation of Bell Crank Lever — This lever is curved, as 
shown in Fig. 112. in order to clear some obstruction. 

185 



Control Systems 



Art. 132 



fri 


?\ z 


1' 


■* 


<' 


•- 


-. — 




-6i 



0<?9l 






1 








X 




f . 





3£CT/0A/ Y-Y 




3ect/oa/ X~X 
Fig. 112. Bell Crank Lever 

Stress on weld between lever and torque tube. 

Minimum area of weld = irX 1.375 X2X-049 = .423 in. 

Moment = 5160/2 = 2580 lbs. 

2580 

Total stress = = 3750 lbs. 

.688 

3750 

f s = = 8880 lbs. per sq. in. (satisfactory) 

.423 

Stress on section X — X., Fig. 112. 

It will be assumed that at this section only half the moment is carried 
by the crank, the rest having been taken up by the half of the weld above 
the section. 

Moment = 2580/2 = 1290 in. lbs. 

Net I = .0430 on section X— X 

1290 X .937 

f b = = 28,100 lbs. per sq. in. 

.0430 



186 



Art. 133 Investigation of Teeth in Collar on Aileron Torque Tube 

55,000 

Coefficient = = 1.96 (satisfactory) 

28,100 

Stress on section Y — Y, Fig. 112. 

I =.0237 M = 397x4.0= 1590 in. lbs 

Z =.0361 P cos 75° = 103 lbs. 
A =.146 

1590 

f b = = 44,000 lbs. per sq. in. 

.0361 

103 700 

P/A = = lbs. per sq. in. 

. H6 44,700 

55,000 

Coefficient = = 1.23 (satisfactory) 

44,700 
Although calculations for the stresses on sections W — W and Z — Z 
are not given, they should be carried through to check the design. 

133. Investigation of Teeth in Collar on Aileron Torque Tube — Fig. 
113 shows the collar and tube. 

Depth of teeth = .25 in. 

Area of one tooth = .25X-125 = .0312 sq. in. 

2580 

Force on each tooth = = 1250 lbs. 

3 X -6875 
1250 

Bearing stress = f c = =40,000 lbs. per sq. in. (satisfactory) 

.0312 

.6875X2X3. 14X- 125 

Section area of one tooth = = .090 sq. in. 

6 
1250 

Shearing stress = f s = =13,900 lbs. per sq. in. (satisfactory) 

.090 

Co//or 




r 

Fig. 113. Teeth on Aileron Torque Tube Collar 

187 



Control Systems 



Art. 134 



134. Investigation of Aileron Torque Tube — Length of the torque 
tube from the end in the fuselage to the center of the ailerons = 120 ins. 

Tube is l T / 4 X 8 / 3 o in. 
Polar I = .1146 

2580 X .625 

Torsional shear = f s = =14,100 lbs. per sq. in. 

.1146 (satisfactory) 

57.2 X 2580 X 120 

Angle of twist = = 12.9° 

12,000,000 X -1146 

12.9X21 

Extra movement of center of hand grip = = 4.7 in. 

57.2 

135. Investigation of Rudder Bar — Owing to the offset in this rud- 
der bar, see Fig. 114, two kinds of stress are produced, bending and 
torsional. 

Moment of rudder load = 2120 in. lbs. 

2120 

Tension in cable = = 472 lbs. 

4.5 

Cable is 1/8 in. 7 X 19 extra flexible of 2000 lbs. capacity. 
2000 

Coefficient = = 4.23 

472 

2120 

Pressure on pedal = = 202 lbs. 

10.5 

Stress on section A — A at which cable is attached. 
Bending moment = 202 X 6= 1212 in. lbs. 
Torsional moment = 202 X 3 = 606 in. lbs. 
Tube is 1 X 1/16 in. 
I = .0203 
Polar I = .0406 



f* = 



f. 



1212 X .5 

.0203 

606 X -5 



.0406 

Maximum combined stresses. 
29,900 
Bending = f b = \- 



= 29,900 lbs. per sq. in. 



7,450 lbs. per sq. in. 



V 



(7450) : 



31,650 lbs. per sq. in. 
188 



(29,900 ) 2 

( 2 ) 

(satisfactory) 



Art. 136 



Investigation of Mast on Elevator Tubes 



Tubc-y 


•M" 


4? 6 


A 


fr^4 


V« 




J^ 


'V 


Judder C(?6/e 
token off here 



Fig. 114. Rudder Bar 



Shearing = f s = 



\/(745 



or- 



(29,900) 



( 2 )■ 

16,700 lbs. per sq. in. (satisfactory) 



136. Investigation of Mast on Elevator Tubes — A foment of elevator 
load = 3620 in. lbs. Refer to Fig. 115. 



3620 



Pull on elevator cable 



4.5 



Stress on weld between coll; 



805 lbs. 



elevator tube and mast. 



3620 
Total stress on weld = = 4130 lbs. 



875 



4130 



Unit stress on weld 



2 X -0625 X 3.14 X 1.75 
= 6030 lbs. per sq. in. (satisfactory) 




Fig. 115. Elevator Control Mast 
189 



Control Systems 



Art. 137 



Stress on section D — D. 

m=1.0X 805 = 805 in. lbs. 

1.20 3 X .40— 1.075 3 X -275 

I = =.0291 

12 
805 X .6 

f b = = 16,600 lbs. per sq. in. (satisfactory) 

.0291 
Stress on section C — C. 

m = 3.6X805 = 2900 in. lbs. 

1.0 X 2. 3 — .875 X 1-875 3 

I = =.185 

12 
2900 X 1.0 

f b = = 15,700 lbs. per sq. in. (satisfactory) 

.185 

137. Attachment of Masts to Tubes — The method of welding used 
in this design to secure control levers and masts to tubes is not at all 
satisfactory. The welding is liable to cause distortion of the tube and 




-Turned 5/eevc 



Fig 116. Method of Securing Large Masts to Torque Tubes 

190 



Art. 137 



Attachment of Masts to Tubes 



lever as well as completely annealing them at the weld. If the gage of 
the metal is light, a considerable part of the steel may be burned. A 
much better type of connection is shown in Fig. 116. The sleeve can be 
cheaply and easily turned. The distance "d" is determined by the 
brazing area required to carry the stress. When the brazing is properly 
done, 25,000 lbs. per sq. in. is a safe ultimate stress. Sometimes the 
sleeve can be brazed to the tube, making a unit construction; or it may 
have to be removable, in which case the sleeve would be suitably pinned 
to the tube. If brazing is employed, the sleeve should be cut off flush 
with the outside of the mast. 

The type of sleeve shown assumes that the mast is stamped in two 
halves which are welded or, preferably, brazed together. Masts or 
levers of this general character are used for exposed work, or on large 
airplanes where the stresses are great and where the overall length of 
the lever is over 15 in. 

For masts or levers to be used inside, and of a length less than 15 in., 
the type of mast and sleeve illustrated in Fig. 117 is well adapted. Such 
construction is used on the control system shown in Fig. 118. It should 



3ect/on /? -/) 




Fig. 117. Small Internal Mast and Its Attachment to Torque Tube 

191 



Control Systems 



Art. 137 




192 



Art. 138 General Comments 

be noted that a flange is put only on the compression side of the mast. 
One point of weakness in this mast, and often in other types, is through 
section B — B. A slight lateral component in the stress in the control 
cable will cause the top of the mast to buckle sideways. 

DUAL STICK CONTROL 

138. General Comments — Fig. 118 illustrates the control system of 
a two-seater pursuit airplane. One of the few questionable features of 
this design for all dual control work is the torque tube, connecting the 
front and rear sticks, by which the aileron torque is transmitted when 
the control is from the rear seat. Unless the workmanship is good, so 
that the clearances can be small, there will be a certain amount of play. 
Also, as shown in the example which follows, an angle of lag is present 
equal to the angle of torsion in the tube between the sticks. In the case 
of this design it amounts to 7 degs. with the full sand load on the four 
ailerons, which is not an objectionable amount. 

The light, duraluminum base for the rudder bar is an efficient design. 
The manner in which it is secured to the floor by six widely spaced 
bolts tends to make the base rigid even with a light flooring. The steel 
strap holding the rudder bar is riveted to the duraluminum base. 

The end thrust bearing for the front aileron torque tube is combined 
with the base of the forward rudder bar. This gives a firm bearing and 
eliminates the necessity for an extra base. 

The mounting of the elevator levers on the torque tube at the base 
of the rear stick gives compactness and simplicity. These levers are 
attached to the tube by flanged collars of the type shown in Fig. 117, 
to which they are brazed. Such a method of connection is to be recom- 
mended. 

139. Investigation of Torque Tube Between Control Sticks — When 
the control is through the rear control stick, the aileron torque must be 
transmitted by the horizontal tube connecting the control sticks. As the 
analysis of this tube presents some new features, it will be given in 
detail. 

Moment of Aileron Load. 

Total moment = Area X average load X mean arm 
= 51.8 X 24 X 7.82 = 9720 in. lbs. 
On each aileron = 2430 in. lbs. 

Since the aileron masts and the aileron lever arms are the same 
length, the moment carried by the aileron torque tube is the same as the 
moment of the aileron loads. 

Tube is \y s X 7r> in. 
Polar I = .1557 

9720 X -6875 

Torsional shear = = f s = = 42,900 lbs. per sq. in. 

1557 (satisfactory) 



Length of tube = 23 ins. 



193 



Control Systems 



Art. 139 



Angle of torsion 



9720X23 X 57.2 
12,000,000X.1557 



6.85' 



In Fig. 119 DE represents the position of the front stick, and BC 
that of the rear stick. ABC, or the angle of lag, equals the torsional 
angle; refer also to Fig. 120. Since point C moves, with reference to 




/3ng/e of Tors/ on or Lag 



Fig. 119. Angle of Lag Between Front and Rear Control Sticks 




Ang/e of Torsion 



Fig. 120. Relation Between Angle of Lag and Torsion 
194 



Art. 140. Torque Tube Carrying Elevator Masts and Rear Stick 



point A, over the arc AC, the tube connecting points D and C must be 
deflected a distance, AC. The force applied at C that is required to do 
this produces bending in member DC. 

Calculation of Deflection of C Due to Bending. 

6.85X4 

BC =- 4.0 ins. AC = = .478 in. = Deflection of C 

57.2 

Calculation of Force Required to Produce Deflection. 

FL 3 3 X 29,000,000 X- 0778 X- 478 

.478 = or F = = 266 lbs. 

3 EI 23. 3 

Calculation of Bending Stress. 

Moment = 266 X 21.5 = 5720 in. lbs. 
5720 X .687 

f b --= = 50,500 lbs. per sq. in. 

.0778 
Calculation of Combined Stresses. 
Maximum bending 

50,500 , /"" ~W00 2 " 

f b = h V 42,900 2 H =75,100 lbs. per sq. in. 

2 2 2 

Maximum shear 

f 50,500T r 

= 49,800 lbs. per sq. in. 

2 J 
Both of these stresses are above the allowable ultimate and hence the 

strength of the tube may be increased by changing to a wall thickness 

of 1/8 in. However, since this rear control for a pursuit airplane is 

really an emergency control, the present design is strong enough for that 

type of airplane. 

140. Torque Tube Carrying Elevator Masts and Rear Stick. 

Moment of Elevator Load. 

Total moment = Area X average load X mean arm 
= 23.0 X 24.0 X 7.65 — 4220 in. lbs. 
The elevator masts and the elevator lever arms on the control stick 
are the same length, so that the moment on the stick equals the moment 
of the elevator loads. 

Calculation of Reaction on Grip of Control Stick. 
Length of stick to centerline of grip = 24^/s in. 
4220 

F = =175 lbs. 

24.12 

195 




Control Systems 



Art. 140 



Tension in Elevator Cables. 
4220 
T = = 422 lbs. 

2 x 5 - . 

Fig. 121 shows this tube with its applied loads. 

Maximum bending moment = 509.5 X 12 — 422 X H 

= 1470 in. lbs. 
Maximum torsional moment = 2110 in. lbs. 
Tube is U/ 4 X -049 in. Polar I — .190 
1470X-875 
f b = : =13,550 lbs. per sq. in. 



.095 
2110X.875 



9,720 lbs. per sq. in. 



.190 

Combined Stresses. 

13,550 

Maximum f b = 1- 

2 



V972 



9720 2 + 



13,500 



18,630 lbs. per sq. in. 



. / f 13,500 



Maximum f s = V 972 ° 2 + 



= 11,850 lbs. per sq .in. 
As far as these stresses are concerned, the tube could be decreased to 
perhaps \y 2 in. diameter. 




Fig. 121. Forces Acting on Rear Stick and Transverse Torque Tube 

196 



Art. 141 Pursuit Airplane Control System 



141. Pursuit Airplane Control System — Fig. 122 illustrates diagram- 
matically the functioning of a control system which has certain good 
points. For the design in question, the aileron and elevator levers are to 
the rear of the seat where there is ample room instead of in a crowded 
cockpit. As shown in Fig. 123, the aileron and elevator levers are com- 
bined in a compact manner. The former is securely riveted to the torque 
tube, and the latter, pinned through the torque tube, rotates within the 
aileron lever in a plane passing through the axis of the tube. The weight 
of the entire system, excluding the torque tube in the wings, is 8 lbs. 

142. Relation of Air Controls to Pilot — It is very difficult to es- 
tablish fixed rules to govern the layout of a standard cockpit because 
of the different types of fuselages found in military airplanes. Yet the 
dimensions directly affecting the pilot's comfort can be standardized. In 
Figs. 125 and 126 are shown three views of a cockpit in which the main 
dimensions are indicated by symbols. Table XXII gives average values 
for these various dimensions, which are based on nine two-seater and 
eight single-seater airplanes of American, British, French and German 
designs. It is recommended that, in general, these values be closely 
adhered to. The angular movements of the rudder bar and control 
stick, and the lengths of the control arms on the elevator and rudder are 
more dependent on the power ratio and linkages desired than on the 
pilot's reach and height. For this reason the values given in the table 
for these angles and dimensions should be regarded merely as reasonable 
values. They will vary with the type of airplane. 

It is very necessary that there should be an adjustment which will 
allow for variation in a pilot's height. The adjustable footpad on the 
rudder bar shown in Fig. 124 is the simplest and most effective method 
of adjustment. The dimension w given in Table XXII is with the foot- 
pad in mid-position. In the case of a very tall pilot, the use of a small 
cushion in the front of the seat adds greatly to his comfort by supporting 
his thigh under the knee. Such a cushion would conform to the curve 
of the seat in plan view and be triangular in section. 

The seat is designed in accordance with the specifications of the 
Parachute Branch, of the Equipment Section of the Engineering Di- 
vision. The maximum dimensions of any parachute pack for which pro- 
vision must be made are 26 in. X 14 in. X 6 in. In most cases the pack 
will be smaller. The 6 in. given as the maximum thickness determines 
the location of the bulkhead just to the rear of the pilot. It is essential 
that the head-rest should not impede a pilot in climbing out of the cock- 
pit with a parachute pack on his back. 

143. Controls for Large Airplanes — For multi-engined bombers and 
commercial airplanes the simple stick control is no longer practicable. 
Some form of wheel control for the ailerons is commonly resorted to. A 
recent design of this type embodying many excellent features is illus- 
trated in Fig. 127. The transmission used is the sprocket-chain combi- 
nation which is mechanically highly efficient, and also simple. Some 
designs employ a worm and gear transmission. With this type, however, 
the loss of power from friction is larger than with the sprocket and chain, 

197 



Control Systems 



Art. 143 




Tube ro E /e rofor Lever 



_<£_ Tor que Totit 




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Fig. 122. Functioning of Control System on VCP-1 



E/cisofor Control 
lever 




Fig. 123. Aileron and Elevator Lever Unit 



198 



Art. 143 



Controls for Large Airplanes 




Braze 




Fig. 124. Adjustable Foot Pad for Rudder Bar 



199 



Control Systems 



Art. 143 



TABLE XXII 







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Fig. 125 

200 



Art. 143 



Controls for Large Airplanes 




Fig. 126. Side Elevation an 



5/DE 1//EUV 

d Plan View of Standard Cockpit 



201 



Control Systems Art. 144 



which is a most important consideration when the power required to 
operate the control must be reduced to a minimum. An improvement 
in the design illustrated would be the use of ball bearings on the sprocket 
and on the two pulleys at the base of the steering column. This would 
be necessary with the largest types of airplanes. The question of the 
power ratio is an important one. In the old "dep" control this ratio 
was about 4 to 1. For heavy airplanes a greater ratio is required, or 
the controls become too sensitive and require too much power to oper- 
ate. The present design has a ratio of 9 to 1, which is ample and insures 
ease of control. A 200-pound pull on the rim of the wheel gives a cable 
tension of 1800 pounds, which is applied to a 12^4 inch aileron mast. 
To give a maximum movement of the ailerons of 25 degs. either side of 
the neutral position requires, with the design shown, 1*4 turns of the 
wheel. The entire weight of this control, including pulleys, guards and 
cables, is less than fifteen pounds. 

Where it is desirable, the horizontal tube at the base of the steering 
column can be extended to the sides of the fuselage and have walking 
beams mounted on it for the elevator cables. This arrangement makes 
the control a unit. 

The fact that a worm and gear control is practically irreversible, is 
one feature much in its favor. For example, if one engine stops in a 
twin-engine airplane the unbalanced torque is partly counter-balanced 
by the ailerons. With a chain and sprocket control a constant force 
must be exerted on the wheel, while in the case of the worm and gear 
type the ailerons, once having been forced to a certain position, will 
tend to remain there, thus relieving the pilot of continued strain. 

144. Ultimate Allowable Stresses — Most of the material used in a 
control system will be either quarter hard carbon steel of a tensile 
strength of 55,000 lbs. per sq. in. or ?>y 2 per cent nickel steel, usually cold 
rolled, of a tensile strength of 100,000 lbs. per sq. in. The low carbon 
material is covered by U.S.A. Signal Corps Specifications 10,201 and 
10,225, the first for sheet stock, the second for tubing. For alloy steel 
either specification 10,029 or 10,045 may be used. All bolts are made 
of this nickel steel, and also certain other members which carry high 
stresses and which would otherwise be unduly heavy. However, if mem- 
bers are made of high strength material it is not permissible to braze 
them, because their strength would be very largely reduced by this 
operation. 

The ultimate strength of all steels in shear may be assumed to be 
very closely 2/3 of its ultimate tensile strength, unless the steel is heat 
treated. This means that the maximum allowable shearing stress on 
bolts may be taken as 65,000 lbs. per sq. in. It should be remembered, 
however, that, on account of necessary clearances, bolts will always be 
subjected to bending stresses in addition to pure shearing stresses. For 
small bolts especially it is not advisable to allow more than 45,000 lbs. 
per sq. in. shear. 

Torsional stresses may be considered the same as shearing stresses. 
The usual formula for torsional stress, f s = M •r/J , will not give the 

202 



Art. 144 



Ultimate Allowable Stresses 



Beiwcen #gfi, the cob/e is' 
rep/oced by P//O &a/dnv/'n 
ro//er chQin. 



^^—/Z x /g M/cke/ J tee/ Tube 




£ &o/ti 



Fig. 127. Wheel Control for Large Airplanes 



true shearing stress when failure in torsion occurs. Upton has developed 
a formula, / s = .75 M r/J, which gives closely the true value of the tor- 
sional stress at failure in solid or thick walled tubes. 

The actual ultimate torsional stress so calculated may be taken as 
equal to 2/3 the ultimate tensile strength. In thin walled tubes sec- 
ondary stresses are produced by local crumpling of the metal so that the 
allowable ultimate stresses are lower than those given above. Fig. 153 
is a curve for this reduction in ultimate torsional shear stress plotted 
against the ratio of diameter to wall thickness. This curve is based on a 
few experimental values and it is believed will give safe and reasonable 
results, but it is subject to change when more data is available. 

Definite recommendations in regard to ultimate bearing strengths 
can not be made. Data on a large number of lug tests indicate that 
ultimate bearing strengths may be taken as nearly twice ultimate tensile 
strengths for any given steel. 

203 



CHAPTER VII 
CONTROL SURFACES 

145. General — For the most part an exact analysis of control sur- 
faces is impossible. The present chapter indicates such methods of 
analysis as are practicable, but its chief aim is to suggest certain features 
of design that tests on control surfaces have shown to add to the strength 
of the surfaces. Perhaps the best means of estimating the strength and 
weight of any given design, and of determining the size of the main mem- 
bers is a tabulation of similar data for a number of surfaces that have 
been tested. Such a tabulation is given in Table XXIII. With few ex- 
ceptions the surfaces tabulated are representative of recent good design. 
Four or five types of construction are each represented by several ex- 
amples. Some of the designs, especially in steel, are somewhat of an 
experimental nature. General problems in design will be taken up first, 
then a discussion of the important features of various designs which are 
illustrated by photographs and sections, followed by an explanation of 
methods of analysis. 

146. Control Masts — Probably the most important single detail of a 
design is the mast and the method of attaching it to the surface. The 
necessity of placing a mast in line with its control cable is clearly illus- 
trated by Fig. 129, in which the angle between the cable and the line 




Fig. 129. USD 9-A Control Mast 

204 



Art. 147 Bracing 

of the mast is 9 degs., giving a side component equal to .16 of the pull 
in the cable. Particularly for the type of mast shown, a core of sheet 
steel with wood fairing, a lateral force causes very severe stresses, owing 
to the small moment of inertia of the section about its long axis. If a 
mast is secured to a rib, or is braced by an external stay wire, these 
also should be in line with the control cable. Of all types of masts, the 
best for exposed work is the hollow, streamline mast of sheet steel, 
welded along the trailing edge, and spread at the base. The mast shown 
in Fig. 149 is an example of this type. A plywood mast with a steel core, 
which is built into a strong box rib that transmits a large amount of 
torque, is illustrated in Fig. 130. Where the height of the mast is small 
this design is entirely satisfactory. If a bracing wire is present, running 
from the mast to the trailing edge or to a stringer rib, the base of the 
mast can be shortened because the torque that the mast carries is much 
reduced. 

The method of securing a mast to a tube by welding to the main 
spar and to a tubular rib, as shown in Fig. 131, can be used when a cir- 
cular tube is employed for the leading edge of the movable surface. 
Since most of such tubes have a very high ratio of diameter to wall thick- 
ness it is always advisable to reinforce the tube, at a point where a con- 
trol mast is attached, by a 3 in. length of 1/16 in. tubing, fitting inside, 
and spot welded or brazed to the main tube. With the type of construc- 
tion in Fig. 129, a steel box, shown in detail in Fig. 132, is a standard 
means of securing a mast to a torque tube. The strip forming the top 
and bottom of the box passes around the tube and is welded to it, and 
also the sides of the box are flanged out and welded to the tube. A rib 
may or may not be fastened to the box. 

147. Bracing — The maximum speed of an airplane is often the de- 
termining factor in deciding whether or not tail surfaces should be ex- 
ternally braced. With velocities over 150 M.P.H. the resistance of ex- 
posed struts and wires becomes so large that it is more economical to 
eliminate them, and to strengthen the spars of the surfaces so that they 
can carry the load as cantilever beams. Internally braced surfaces will 
thus be confined almost wholly to high speed pursuit airplanes which 
have surfaces of a relatively short span and of a roughly semi-circular 
or elliptical shape. As will be noted in describing the different types of 
surfaces, the surface with a fairly thick section and with wooden spars 
is best adapted to internal bracing. As the possibility of doing away 
with external bracing depends not only on the strength and rigidity of 
the control surfaces themselves, but also on the character of the support 
afforded by the fuselage, this character must be considered. A fuselage 
that is narrow at the end is unsuited to an internally braced empennage. 

Type 1. Fig. 130 is a general view of this tail unit, showing the in- 
ternal construction of the horizontal surfaces, and the character of the 
fuselage. Reference to Table XXIII shows that although these surfaces 
are unbraced their unit weight is reasonably low. The efficient design 
which is the cause of this low weight makes a careful study of the details 
worth while. 

205 



Control Surfaces 



Art. 147 



















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206 



Art. 147 



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207 



Art. 147 



Control Surfaces 




Art. 147 



Bracing 




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4) 



209 



Control Surfaces 



Art. 147 




Fig. 132. Steel Box for Supporting Control Mast 



The heavy mast ribs carry a considerable portion of the torque into 
an auxiliary spar. Owing to the fact that the elevator is a single unit 
this spar is continuous and is, therefore, peculiarly well fitted to transmit 
loads to the ribs. This condition will not occur with a double elevator 
unless each mast is located nearly at the center of its elevator. The 
auxiliary spar performs two functions; it reduces the torque carried by 
the main spar, and it supports the ribs, as well as bracing them laterally. 
A section of this member is shown in Fig. 133. Since the moment re- 
sisted by this secondary spar decreases from the center, the width of its 
flange could be reduced and circular lightening holes cut in the web 
toward the ends. 

A section of the mast rib is shown in Fig. 134. It will be noticed that 
this rib is of a box section out to the cantilever spar. The balance of 
the ribs are of light construction with 7/16 X 1/8 in. spruce capstrips 
and 3/32 in. 3-ply webs. The manner in which the cap strips are at- 
tached to the main spar is worth attention. The large surface provided 
for gluing and nailing insures a rigid connection. Fig. 135 is a section 
of the heavy, stabilizer box rib. The short stabilizer ribs are similar to 
the light elevator ribs, while the transverse rib is of sturdier construction 
with 1/8 in. web and 1/2 X 1/8 m - ca P strips. All plywood, both on 
the spars and ribs, is 3-ply spruce. The leading and trailing edges of 

210 



Art. 147 



Bracing 




& ^^^ 



<3 P/y fenee/- 



./S 



'Spruce 



Fig. 133. Auxiliary Spar in Orenco 
Elevator 



$>P/y 



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6 



Fig. 134. Section of Mast Rib in 
Orenco Elevator 



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3 P/y L/enee 



a 



"^ r?o/~>Ogony 



S 
'32 



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Fig. 135. Section of Box Rib in Orenco Stabilizer Next to Fuselage 



211 



Control Surfaces 



Art. 147 



stabilizer and elevator, respectively, are 7/16 in. O.D. .035 in. gage 
aluminum tubing, fastened to the ribs by means of a copper strap 
wrapped around the tube and nailed to the rib. 

The front spar of the stabilizer is spruce of rectangular section 
2 11 /ie x Va m - For economy this spar should decrease in width from 
the body out. In Fig. 136 is shown a section of the rear stabilizer spar, 
and in Fig. 137, the forward elevator spar. The box construction of the 
latter has been followed in most of the best recent designs. The wide 
spar is strong in torsion and provides a broad base to which elevator 
masts can be secured. At hinge and mast points the routing is omitted. 

The construction of the rudder is made clear in Fig. 138. The some- 
what excessive weight of the rudder is due to its excess strength. Aside 
from the two light wood ribs, the rudder is all steel. The torque tube 
is 2 in. O.D. and .065 in. gage; all other tubing is 7/16 in. O.D. and .035 
in. gage. These gages can be reduced to at least .042 in. and .028 in., re- 
spectively, and in the case of the 7/16 in. tubing and the metal ribs 
perhaps to .022 in., changes which would bring the weight of the rudder 
down to about 1.00 lb. per sq. ft. 

The Orenco Type D fuselage, tapering down to a horizontal edge 
in the rear, is well suited to internally braced horizontal surfaces be- 
cause of the broad, firm support afforded the spars. Conditions are not 
quite as favorable for the rudder, which had to be located rather far 
forward on the fuselage in order to get sufficient depth in the latter to 
give adequate bracing. The chief effect of this is to place the rudder 
area too high above the thrust line. Taken as a whole, however, the 
design of the empennage is excellent. 




ESI 



Fig. 136. Rear Spar in Orenco Stabilizer 

212 



Art. 147 



Bracing 



Top 




Fig. 137. Leading Edge of Orenco Elevator 




Fig. 138. Internal Construction of Orenco Rudder 



213 



Control Surfaces Art. 147 

Type 2. The tail surfaces of the Curtiss Kirkham 18-T, shown in 
Fig. 139, are of the same general type as those just discussed, but as 
they present considerable differences in detail they will be taken up here. 
The original manner in which, with a. fuselage of rather narrow elliptical 
section, a sufficiently rigid support is obtained for the unbraced sur- 
faces should be noted. The fuselage is cut off short and the blunt end 
streamlined by false fairing. As will be seen, all controls are internal 
so that this design practically eliminates structural resistance. 

Both fixed tail surfaces are entirely covered with 1/16 in., 3-ply 
Spanish cedar plywood, covered with linen, doped and painted. Mov- 
able surfaces have ribs of pressed steel, similar to those shown in Fig. 
138, welded or brazed to a steel torque tube. The leading edge of the 
fin and stabilizer is a small spruce member of triangular section, and 
the rudder and elevator trailing edges are of .022 in. gage steel bent to 
a U section. The ribs in the fixed surfaces are spaced 9 in., and are of 
the type shown in Fig. 140. The ribs in the movable surfaces are 
laterally braced by light wood spacers of rectangular section 3/8 x 5/8 
in. One is used in the elevator and two in the rudder. 

The construction of the rudder is evident from Fig. 141. One fea- 
ture deserves attention, the close spacing of the ribs, 6 in. This is made 
necessary by the unusually great width of the rudder which causes the 
moment on each rib near the torque tube to be large. As a general rule, 
the ribs in control surfaces are spaced too widely. Since the ribs are such 




I 



:00 



Fig. 139. Empennage of Curtiss Kirkham 18-T Triplane 
214 



Art. 147 



Bracing 



•3pruce 



^NH 



For 

5tob///zer j£~* 

for Fin % - 

l/ert/co/ 3prc/ce 
J/r/& — - 

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a 



Fig. 140. Stabilizer Rib on Curtiss Kirkham 



"curtiss Kmmvf 

TFIPLANt 



»~Ojr .».»/» ft 




Fig. 141. Internal Construction of Curtiss Kirkham 18-T Rudder 

215 



Control Surfaces 



Art. 147 



a small part of the weight of a surface, the addition of two or three of 
them will often considerably increase its strength and rigidity without 
adding appreciably to the weight. The diagonal tube running up from 
the torque tube to the upper edge of the rudder serves to prevent ex- 
cessive deflection of this portion of the rudder. 




Fig. 142. Section of Stabilizer Spar on Curtiss Kirkham Half Way Out 




Fig. 143. Maximum Section of Stabilizer Spar on Curtiss Kirkham 18-T 



216 



Art. 147 



Bracing 



The entire load on the horizontal and vertical surfaces is carried in 
bending by the spars in the fixed surfaces acting as cantilevers in com- 
bination with the torque tubes in the elevator and rudder. The stabilizer 
and fin spars increase in section toward the fuselage. Fig. 142 is a sec- 
tion through the stabilizer spar half way out, and Fig. 143 the maximum 
section at the fuselage. The fin spar is of the same type, but slightly 
smaller. For the torque tubes on rudder and elevator, \y 2 in. O.D. 
tubing was employed, of .035 in. gage for the elevator and .032 in. gage 
for the rudder. 

The hinge used on these surfaces has some excellent features. Fig. 
139 shows it to be of the streamline type that allows very little leakage. 
Fig. 144 illustrates the design in detail. The guide collar and the hinge 
must be slipped on the torque tube before the ribs and frame are welded 
to it. Aside from the great difficulty of replacing a broken strap, this 
hinge is probably the simplest and best kind in use. Fig. 145 illustrates 
a hinge that overcomes this objection. 

Type 3. The type of surface illustrated in Fig. 146 is well suited to 
good sized surfaces on two-seater airplanes. The construction is very 
simple and strong. Although the ultimate strength of this design is 
given in Table XXIII as only 20 lbs. per sq. ft., neither the elevator nor 
stabilizer was injured at this load, but the test was discontinued on ac- 
count of the complete failure of the control system. The weight of the 




3"azec/ 



)Or 3 ' 



3u/des 



Fig. 144. Elevator Hinge on Curtiss Kirkham 18-T 



217 



Control Surfaces 



Art. 147 



ribs may be decreased by using 3/32 in. 3-ply spruce lightened by cir- 
cular holes. The secondary spar is a 1/2 in. spruce rod. A strip of 
.065 in. gage aluminum forms the trailing edge, with bent ash for the 
curved portion at the tips. The main spar is constructed of spruce and 
3-ply mahogany. Its width is 6 in. As is evident in Fig. 146, the spar 
offers a broad base for the control masts, and is also capable of carrying 
a large torque. Under the masts, the spar should be reinforced by 
blocks of hardwood, about 2 in. long. This will prevent a crushing of 
the spruce web at the point of attachment of the mast. These webs 
may be routed except at mast and strut points, and tapered slightly in 
plan view to advantage. The horizontal surfaces are braced by two 
11/16 in. O.D., .049 in. gage struts running from the bottom of the 
fuselage, one to the end of the rear stabilizer spar and one to the forward 
stabilizer spar at a point 3/5 way out from the fuselage. 

Type 4. The thin surface of tubular construction illustrated in Fig. 
147 has been used to a considerable extent. From the aerodynamic 
point of view these surfaces are not as efficient as the thicker, double- 
cambered surfaces, i.e., their L/D is less and also their maximum K y 
or unit lift. One point in their favor is the rapidity with which the lift 
builds up. They are more sensitive than thicker sections; in other 
words, a less angle of incidence is required to give the needed force. 
Thin surfaces are perhaps better adapted to slow or moderate speed 
airplanes than to pursuit airplanes, because the high speed of the latter 
insures sensitive controls with either type of surface. Furthermore, as 
it is desirable to eliminate external bracing in fast airplanes, thin sur- 
faces, which must be braced, are not suitable' for such airplanes. The 
simplicity of the tubular construction makes it good production work. 

One of the principal problems in tubular work is the joint between 
the torque tube and the ribs. Formerly, this was made by slotting each 
member, inserting a gusset plate, and brazing the parts together. Slot- 




Fig. 145. Elevator Hinge with Removable Strap 

218 



Art. 147 



B 



racing 










K 



219 



Control Surfaces 



Art. 147 



ting the main tube and rib weakened them materially. A much more 
satisfactory joint is shown in Fig. 148. It will be observed from Fig. 147 
that where a rib of smaller diameter than the torque tube is joined to 
the latter the rib is flattened so that its long axis is vertical, while at the 
junction of the rib and trailing edge the rib is flattened so that its long 
axis is horizonal. 




Fig. 147. Rudder of Tubular Construction 

Conne c f/on Str/p ■ 




Span 

— We/cf- 

Fig. 148. Method of Joining Ribs and Main Spars 

220 



Art. 147 Bracing 

Owing to their low section modulus, neither the spar nor rib tubes 
can carry large bending moments. It is, therefore, necessary to brace 
the main spar from the fuselage, and the trailing edge or auxiliary spar 
from the mast. For this reason,' the mast rib, and also the trailing edge 
or secondary spar, should be fairly stout. The importance of putting 
the mast and mast rib in line with the control cable should again be em- 
phasized. It is well to locate hinges as near the ends of the elevator or 
rudder torque tubes as possible to avoid excessive deflections of the tips. 
If a tip cannot be properly braced by ribs at right angles to the torque 
tube, a diagonal bracing tube is run from the latter out to the end of 
the tip, as in Fig. 141. 

Because of the large excess strength of the Curtiss JN-6 HB rudder, 
its weight can be cut materially by reducing the gage of the torque tube 
and mast rib from .049 in. and the gage of the other ribs from .035 in. 
The gage of the trailing edge and of the auxiliary spar should not be 
decreased. In fact, since with this type of surface the auxiliary spar 
largely increases the strength and rigidity of the structure, it might be 
well to increase slightly the diameter of this member which is now 
7/16 in. In changing to a smaller gage it should be remembered that 
the ultimate unit strength of the metal is reduced. 

Type 5. A design which retains most of the desirable features of 
the steel, tubular construction, but which has the aerodynamic efficiency 
of the double-cambered surface, is illustrated in Fig. 131. The wood 
ribs are secured to the torque tube and trailing edge by means of light 
straps welded to the tubes and nailed to the ribs. The heavy mast rib 
and secondary spar should be noted. 

Type 6. For the thick double-cambered surfaces of large two- 
seater airplanes, the general construction shown in Fig. 149 has proved 
very efficient. The ribs and spars are of light gage, pressed steel. All 
joints are welded. A detail sketch of the mast box is shown in Fig. 132. 
The manner of bracing the stabilizer with a horizontal truss may be 
noted. For the compression struts 1/2 in. tubing is used. Instead of a 
tube, a composite spar made up of a steel member, similar to the front 
stabilizer spar, and a spruce member is employed for the stabilizer 
trailing edge. The construction of the hinge is evident from the general 
photograph. It is noted that the width of the hinge is greater than 
necessary. A wide hinge strap produces excessive friction. An excellent 
example of this type of hinge is illustrated in Fig. 150. In passing, 
attention should be called to the wire or light strap, welded to the 
elevator tube, which forms a loop about the hinge strap. The fabric 
on the elevator is sewed to this loop. The surfaces are externally 
braced by wires from the rudder post and fuselage to the end of the 
stabilizer spar. 

In this construction the elevator ribs are of deep enough section at 
the torque tube to act efficiently as cantilevers with no support from the 
trailing edge or a secondary spar. It is, therefore, more economical 
to omit an auxiliary spar such as is shown in Fig. 129 and brace the ribs 
laterally with piano wire, spot welded to their chords. The weight thus 

221 



Control Surfaces 



Art. 147 




Fig. 149. Truss Type of All-Steel Control Surface Construction 




Fig. 150. Aileron Hinge 

222 



Art. 148 Loading 

saved should go into extra ribs. In the test, the first failure occurred 
by buckling of the chords of the elevator ribs. As a rule, the spacing 
of ribs should not exceed 12 in. Final failure took place through shear- 
ing of the bolts securing the lug of the brace wire from the fin. As can 
be seen from the photograph the diagonals on the stabilizer ribs are too 
slender. A slight increase in their section adds very slightly to the total 
weight. A good value for the gage of all ribs is .020 in. ' This is suf- 
ficiently heavy, and if lighter material is used the construction is not 
stiff enough. As the ribs form only a small portion of the total weight, 
it is not worth while to cut down their gage too much. 

When a tube is used for the trailing edge of a stabilizer the type of 
hinge shown in Fig. 151 has given satisfaction. With this kind of hinge 
the strap around the elevator tube can be easily replaced. In con- 
junction with this hinge a light support, or former, Fig. 152, to which the 
fabric is attached, is secured to the stabilizer tube every 6 in. by little 
legs. This arrangement reduces the leakage and streamlines the hinge. 

For main spars which are subjected to torsional stresses in excess of 
10,000 lbs. per sq. in., and preferably for all spars of steel tubing, seam- 
less tubes must be used instead of "locked seanv tubes which are de- 
cidedlv weak in torsion. 



< qe St rap -y 

Stab///zer 
Tube 



Le 9^P 





fiors to au/de Strap 
Fig. 151. Aileron Hinge Fig. 152 

STRESS ANALYSIS 

148. Loading — The loading which is made the basis of an analysis 
should be that given in the Government Specification for tail surfaces. 
The distribution of load between the fixed and movable surfaces is 
shown in Fig. 107. its intensity for the different types of airplanes being 
stated in the Specifications. 

149. Unbraced Surfaces — The calculation of the external moments 
on an unbraced empennage is a simple, determinate problem. In most 
cases there is no front stabilizer spar, and the entire bending moment is 

223 



Control Surfaces Art. 150 



carried by the rear stabilizer spar acting with the elevator spar. In case 
auxiliary elevator or stabilizer spars are present, their effect in relieving 
the main spars may be neglected, or some arbitrary allowance made for 
them. The moment of the loads on both surfaces should first be cal- 
culated about the outer point of support of these spars. It may be 
noted that in this computation the same moment will be obtained if the 
load per sq. ft. on the elevator is taken as uniform instead of triangular. 
This moment, divided by the ultimate strength of the material, gives the 
total necessary section modulus for both spars. In case the stabilizer 
spar is wood, and the elevator torque tube steel, the proportion of the 
total load carried by each member must be determined by the method 
explained in Art. 25. The principle upon which this method is based 
is that both spars deflect together. When the moment resisted by each 
spar is thus computed the member is designed as before. In addition to 
bending stresses, the elevator torque tube is subjected to torsional 
stresses, which may be calculated by the formulas of Art. 26. The tor- 
sional stresses so obtained should then be combined with the bending 
stress according to the formulas given in the above article. For wood 
members the strength of the material in horizontal rather than in 
direct shear will probably be the limiting factor. In order to secure 
lightness, spars as a rule taper toward the outer end. It is well to check 
the stresses at one or more points to guard against excessive taper. If 
steel tubes are employed for the members, especially where the tubes 
are long and where the torsional stresses are high, a reduction in weight 
may be secured by splicing the tubes as shown in Fig. 154, using for the 
outer section a tube of the same diameter as the main section, but of 
lighter gage. Splicing should be resorted to for special cases only. The 
ultimate stresses for torsion given in Fig. 153 should not be exceeded. 
This curve is an arbitrary one based on a limited number of tests, but it 
will give reasonable and safe results. Ratios of D/t greater than 125 
should never be used, and a ratio of 100 is a better limiting value. The 
same methods of analysis apply to the fin and rudder unit also. 

150. Ribs — Except for steel ribs which are in the form of a truss or 
tube, no accurate analysis can be made of the stresses in a rib. For in- 
determinate cases the design must be purely empirical and based on 
good practice as outlined in Art. 83. With a truss rib the usual assump- 
tions are made that the members are pin jointed and carry only direct 
stresses. The distributed load on the rib must be divided up into con- 
centrated loads applied at the panel points. The graphical method of 
solution explained in Art. 12 will be found convenient. The design of 
compression members in the rib can be based on the parabolic column 
formula of Art. 24. In selecting the proper value the members should 
be assumed to be hinged at the ends. Where a rib is continuous over 
a spar, as in the case of a stabilizer rib supported by a front stabilizer 
spar, the best method of solution is to assume a diagonal in place of 
the spar. The reaction of the rib on the spar may be considered as two 
concentrated loads acting at the panel points adjacent to the spar. 
With tubular ribs the analysis and design is simple. The maximum 

224 



Art. 150 



Ribs 



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Fig. 154. Splice in Tubular Spar 



stress on a cantilever rib occurs at the torque tube, and equals the 
moment of the distributed load carried by the rib, divided by the section 
modulus of the tube at that point. A mast rib which transmits torque 
from the mast to the trailing edge or to the auxiliary spar of a surface is 
an indeterminate structure, whether the mast is braced or not. If a gin- 
wire is used, the rib is subjected to a heavy compression equal to the 
horizontal component of the stress in the wire, and to a bending moment 
equal to the algebraic sum of the moments of the vertical component in 
the wire, the distributed load on the rib, and the reaction of the auxiliary 
spar and trailing edge. If a secondary spar is present the worst stress in 

225 



Control Surfaces 



Art. 151 



the rib probably occurs at the point at which the reaction from this spar 
goes into the rib. When the mast is unbraced the rib carries only bend- 
ing moment. An estimate of the reactions of a secondary spar at the 
mast rib can be only approximate as they are dependent upon the rel- 
ative deflections of the secondary spar or trailing edge and the various 
ribs. These deflections in turn vary with the width and length of the 
surface, as well as with the moment of inertia of the ribs, auxiliary spar, 
trailing edge, and main torque tube. The angular deflection of the last 
member affects the deflection of the ribs. Another important factor in 
the problem is the location of the mast rib. If, for instance, this rib 
should be placed near the inner end of the surface, a much larger pro- 
portion of the torque from the mast would be transmitted directly to the 
ribs by the main torque tube instead of going into the mast rib. 

151. Braced Surfaces — When a surface is braced by stay wires from 
the fuselage and fin to the stabilizer spar, it becomes a statically indeter- 
minate structure if the spars are continuous over the fuselage. Should the 
spars be hinged at the fuselage the vertical component of the stress in 
the stay wire equals the moment of the distributed load on the stabilizer 
and elevator about the point at which the spars are hinged, divided by 
the distance between this hinge point and the point of attachment of the 
wire. With the spars continuous, this vertical component of the wire 
stress equals the reaction at the point of attachment of the wire which 
would be obtained from a solution of the "three moment equation" 
written for the entire length of the spars. Owing to the fact that the 
horizontal surfaces are symmetrical about the centerline of the airplane 
a very simple expression can be written for this reaction. Referring to 
Fig. 155, R ± = —zv (.375 L x -j- L + .75 L^/L ± ), in which w is the total 



^ 


I / 




e\ 




i 


*/ 














Fig. 155. Example for Solution of Braced Surfaces 

226 






Art. 151 Braced Surfaces 

load per in. run on the surfaces. The rounding of the ends of the sur- 
faces is allowed for by decreasing the span so that the shaded areas in 
Fig. 155 are equal, an approximation that is permissible in most cases. 
The assumption that the spars are supported at a single point at the 
fuselage, instead of at two points as is usual, introduces a slight error. 
The fact that the stress in the wire produces in it a certain amount of 
elongation, thus allowing point 1 to fall slightly below the level of point 
2, is another source of error. To neglect this increases the stress in 
the wire about 3 or 4 per cent. The effect of this deflection can be cal- 
culated, if desired, by using the '"three moment equation," 2m 1 L l -f- 
4 m.,L l = wL^/2 -f- 12 EIz> l /L 1 , in which z>, is the deflection of point 1 
with reference to point 2, to recompute the bending moment at 2. and 
from that to correct the reaction. 

The horizontal component of the wire stress equals the compression 
that is put into the stabilizer spar which may be assumed to carry the 
entire compression. The center of the span is probably the critical sec- 
tion for the stabilizer spar and should be investigated for combined bend- 
ing and compression. 

In determining the stresses in a braced rudder post the method of 
deflection is perhaps best. It consists of computing the deflection due 
to a uniform load of the cantilever rudder post, at the point of attach- 
ment of the brace wire, as if no wire were present. Then the concen- 
trated load, applied at this same point and perpendicular to the axis of 
the rudder post, which would be required to produce an equal and 
opposite deflection, is calculated. This load equals the component in 
the stay wire perpendicular to the rudder post. For a uniformly loaded 
cantilever beam the deflection at any point equals 

w f x 4 L :; x L 4 
d = 



2 EI 112 13 4 

x being measured from the outer limit of loading, and L being the loaded 
length. For a cantilever beam with a concentrated load applied at a dis- 
tance L A from the support, the deflection under the load equals 
d= WLy/Z EI. These two deflections are then equated and a solution 
made for W. Since the value of EI is common to both formulas it may 
be cancelled for the deflection. 



227 



CHAPTER VIII 
FUSELAGE 

152. Truss Type — Fuselages may be classified under three main 
types: Truss, Monocoque and Semi-monoooque. The truss type con- 
sists of a box frame made up of four longerons, running the length of the 
fuselage, connected by vertical and horizontal members so as to form 
four trusses. Truss fuselages may be divided into two sub-types; one 
in which the sides of the fuselage are of cloth and serve only to decrease 
the air resistance, another in which the sides are of plywood and rein- 
force the trusses. The first sub-type is frequently known as the "stick 
and wire" type, as most fuselages built in this manner consist of Pratt 
trusses with wooden longerons and struts, and steel wire diagonals. The 
forward portion of the fuselage, however, is often constructed as a War- 
ren truss. The name stick and wire is not precise since some fuselages 
of this type, for example the Fokker D-7, are of metal construction. 

The chief advantage of the stick and wire fuselage is the ease with 
which it can be repaired. The chief disadvantages are: the continual ad- 
justment or truing up necessary to maintain the fuselage in condition 
for flying; the complexity of the structure owing to the large number of 
tie-rods and fittings; and its "vulnerability to attack" due to the fact 
that the failure of any one member results in serious crippling of the 
fuselage. The first and second disadvantages may be largely overcome, 
as in the Fokker D-7, by using a welded metal framework. The third 
is of importance only in certain types of military airplanes. 

In the plywood covered type the diagonals and struts are wood mem- 
bers glued and nailed to the plywood sides, which act very much as 
gusset plates. This construction eliminates the numerous fittings found 
in the type previously discussed. It has the further advantages of being 
cheaper and easier to build, requiring no "truing up," and being less vul- 
nerable to attack. It is, however, somewhat more difficult to repair 
than the stick and wire type. 

153. Monocoque Type — The distinguishing feature of this type is 
the single-piece shell of plywood which forms the fuselage. This shell 
is built up in a mold which is removed after the shell has been con- 
structed. After the removal of the mold the shell is usually rein- 
forced by bulkheads. The building of the shell is carried on in different 
ways by various manufacturers. One firm which makes a specialty of 
this type of construction uses the following method. A collapsible form 
is built up and covered with unbleached muslin. The first layer of 
veneer strips is then lightly tacked on and held firmly in place by straps 
and special rolls. These rolls are so designed that a single strip of 
veneer can be taken off at a time. Each strip is taken off the mold, 
coated with glue and replaced. The next layer is of unbleached mus- 
lin tape, which is followed, in turn, by two layers of veneer and the 
outside layer of unbleached muslin. The veneer is in long strips 2 to 3 
in. wide and usually 1/16 in. thick. The first layer is wound in a spiral 
making an angle of about 30 degs. with the axis of the fuselage. The 

228 



Art. 154 Semi-Monocoque 

second layer is also in a spiral but in the opposite direction. The strips 
of the third layer are laid parallel to the axis of the fuselage. 

The main advantages of the monocoque type are: its combination of 
lightness and strength; its strength under attack, in that small injuries 
do not render the fuselage unsafe; and the ease with which it can be 
made with a good streamline shape. The most serious disadvantage of 
this type is the difficulty experienced in making satisfactory repairs. 
Other disadvantages urged against it are that unusual skill and excessive 
time are required in its construction and that it is much more costly. 
The mold is expensive, but in quantity production its cost is divided up, 
so that very little need be charged for it against any one fuselage, making 
it a serious disadvantage only in experimental work where but a few 
fuselages of a given design are to be made. The disadvantages of the 
skill and time required have already been reduced by proper methods 
of construction so that this type of fuselage can be constructed as cheaply 
and quickly as the truss type when it is produced in quantity. 

154. Semi-Monocoque — In this type the shell is reinforced by bulk- 
heads and longerons, the former being spaced at intervals of about X 1 /^ 
to 2 ft. In large fuselages of this type the longerons are assumed to 
carry all of the direct stresses and the plywood shell to bind them to- 
gether so that they will act as a unit and carry the shear. In smaller 
examples the shell is assumed to carry some of the direct stresses also. 
The longerons are usually of spruce or ash; the plywood of spruce, pop- 
lar, redwood, or mahogany; and the bulkheads of metal or plywood. 

155. Airplane Considered for Analysis and Loading Conditions — 
T l ie fuselage of a Vought YE-7 two-seater training biplane will be taken 
for analysis. This fuselage is ol the truss or "stick and wire" type and 
is standard. 

Four conditions of loading are considered: 

1. Flying Condition with Elevator Up — In this condition the fol- 
lowing forces are acting; forces due to the weight of the airplane 
and its equipment; propeller forces, torque and traction; and the 
air load on the horizontal tail surfaces. 

2. Flying Condition with Rudder Turned — The only forces con- 
sidered are the air loads on the vertical tail surfaces and the 
reactions at the wings. 

3. Landing with Tail Up — The propeller axis is considered hori- 
zontal, and the forces acting are the weight of the airplane and its 
equipment, the reactions from the chassis struts, and a balancing 
air load on the tail surfaces. 

4. Three-Point Landing — In this condition the air load on the 
tail surfaces is replaced by the reaction at the tail skid. 

In all these cases the weight of the wings is omitted as it is not car- 
ried by the fuselage in flying and not after the airplane has touched the 
ground, until it has slowed down considerably. 

When the critical loads occur, the effect of the weight of the different 
parts of the airplane is increased by the fact that the airplane has a large 

229 



Fuselage Art. 156 



angular acceleration or is subject to shock. To take care of this dynamic 
effect and also for defects of workmanship and material the dead loads 
must be multiplied by a factor called sometimes a load factor and some- 
times a dynamic factor, though the former term is preferable. For air 
loads 5 lbs. per sq. ft. is considered the equivalent of a load factor of 
one. If a fuselage is designed for a load factor of 7, the ultimate air 
load will be considered equal to 35 lbs. per sq. ft. 

156. Computation of Loads for Flying Conditions — The loads due 
to the weight of the airplane and its equipment, their location, and their 
center of gravity are shown in the balance drawing, Fig. 156. Table 
XXIV gives the same information in tabular form. 



TABLE XXIV 
BALANCE TABLE 

Moment Arm 
Item of Group Weight About Rear Face Moment 

Propeller Hub 
(Lbs.) (In.) (In. Lbs.) 

Propeller 30 —3.0 —90 

Radiator. Water and Piping.. 100 5.0 500 

Engine 520 . 21.0 10,920 

Oil Tank and Oil Radiator, 

30 lbs. Oil 45 27.0 1,215 

Gas Tank and Gas (Front) ... 130 50.0 6,650 

Chassis 95 52.0 4,940 

Passenger 170 74.0 12,580 

Gas Tank and Gas (Rear) .... 75 101.0 7,575 

Pilot 170 110.0 18,700 

Tail Skid 10 236.0 2,360 

Empennage 50 256.0 12,800 

Fuselage (inc. Inst., Controls, 

Cowl., Seats, Piping, Wiring, 

Etc.) 335 69.0 23,115 

Wing Cell 305 6v0 19,825 

Total 2035 59.5 121,090 

Without Wing 1730 58.5 101,265 

Fig. 157 indicates the loads and their position relative to the vertical 
fuselage trusses as taken from the balance drawing. In computing the 
stresses in the fuselage it is considered as composed of two trusses loaded 
at the panel points. Fig. 158 shows one of these trusses with the panel 
loads as computed from the balance drawing. The computations to ob- 
tain these panel loads are as follows: 

230 



Art. 156 



Computations of Loads for Flying Conditions 




Fuselage 



Art. 156 



ill 




Fig. 157. Load Positions Relative to Fuselage Trusses 



I'gJV/'/sy /?eoctiOnS2^M 




Fig. 158. Panel Loads Due to Dead Weight 



Station Station 1 Station 2 

Engine 1 Engine 1 Engine ) 

Radiator }. . . . 379 Radiator [.... 197 Radiator L... 74 
Propeller J Propeller J Propeller J 

Fuselage 45 Oil and tank Oil and tank 

11 5 

For two trusses 424 45 X — 31 45 X — 14 

For one truss.. 212 16 16 

Fuselage 50 Gas and tank 

7 

For two trusses 278 130 X — .".'..'. 48 
For one truss. . 139 19 

Chassis 
34 

95 X — 67 

48 
Fuselage 75 

For two trusses 278 
For one truss. . 139 
232 



Art. 156 



Computations of Loads for Flying Conditions 



Station 3 

Gas and tank 

12 
130 X— 82 

19 
Passenger 

12 
170 X— 71 

29 
Fuselage 51 

For two trusses 204 



For one truss 



102 



Station 4 

Chassis 

14 
93 X — 28 

48 
Passenger 

17 
170 X— 99 

29 
Gas and tank 

18 
75 X- 41 

33 
Pilot 

9 
170 X— 46 

33 
Fuselage 40 

For two trusses 254 
For one truss. . 127 



Station 5 

Gas and tank 

15 
75 X — 34 

33 
Pilot 

24 
170 X — 124 

33 
Fuselage 30 

For two trusses 188 
For one truss . . 94 



Station 9 

Tail skid 
18 

10 X— 

20 
Fuselage 

For two trusses 
For one truss. . 



Station 10 

Tail skid 
2 

9 10 X— 

20 
8 Empennage . . . 

— Fuselage 

17 

8 For two trusses 
For one truss. . 



50 
6 

57 
29 



At stations 6, 7, and 8 a load of 5 lbs. on each truss is taken for the 
weight of the fuselage. 

The loads should be divided between the panel points in such a man- 
ner that the center of gravity of the panel loads will be the same as the 
center of gravity of the loads considered acting. In order to obtain this 
result most of the concentrated loads should be divided between the 
two adjacent panel points in inverse proportion to the distances from 
the loads to the panel points. The weight of the fuselage and the engine 
and any weights in front of the first panel point cannot be treated in 
this manner. The weight of the fuselage should be distributed among 
all the panel points in such a manner as to have the center of gravity 
of the panel weights coincident with the center of gravity of the fuselage. 
This can be done only by trial. When the center of gravity of the 
fuselage is unknown it can be assumed to be between 40 and 45 per 

233 



Fuselage Art. 157 

cent to the rear of the forward end. Approximately 60 per cent of the 
weight of the fuselage will be forward of its center of gravity and the 
remaining 40 per cent to the rear. The quickest way to distribute the 
fuselage load is to assume all the panel loads but two, and compute 
these two, knowing the center of gravity and the sum of the two un- 
known loads. If the values obtained are reasonable they should be 
accepted. If they are not, another trial should be made. 

Where a radial or rotary engine is used and is fixed to an engine 
plate forming the front of the fuselage, the weight of the engine and 
propeller unit can be cared for by replacing it with a down load and a 
couple at the first panel point of the fuselage. In most cases the power 
unit is carried on a pair of engine bearers supported at two or more 
points. When the bearers are supported at only two points they can 
be computed as simply supported beams and the reactions easily found. 
If they are supported at three or more points they should be treated as 
continuous beams. Between adjacent supports the load due to the en- 
gine may be considered uniformly distributed, and the load per inch 
run in each span should be such that the location of the center of gravity 
of the engine remains unchanged. The propeller may be considered as 
a concentrated load on the cantilever end of the engine bearers. The re- 
action at each of the supports of the engine bearers can then be com- 
puted. 

157. Air Loads on Horizontal Tail Surfaces — At Station 10 an air 
load of 5 lbs. per sq. ft. should be assumed. This value represents the 
increase of tail load occurring when the angular acceleration of the air- 
plane is increased so that the dynamic effect of each component mass in 
the airplane is increased by the weight of that mass. The maximum tail 
load occurs while pulling out of a dive and may reach very high values. 
Instead of trying to design the fuselage and tail to carry the maximum 
possible values of the tail load, they are designed to be only as strong as 
the wings. A discussion of the strength required of the wings in the 
diving condition can be found in Art. 40. 

There are several methods of calculating the tail load for which the 
fuselage should be designed in order to have the same strength as the 
wings. Three of these will be mentioned. U. S. Air Service Specifica- 
tion No. 1003 gives the formula, F'= 0.0026 AV 2 , where F is the force 
on the tail, A is the area of the horizontal tail surfaces, and V is the nor- 
mal high speed of the airplane. For the case under consideration the 
formula becomes: F = 0.0026 X 32.7 X 108 2 = 993 lbs. The intensity 
of this load will be 993/32.7 = 30.3 lbs. per sq. ft. 

A British Admiralty report gives the formula, F=\.SWC/L, 
where W is the weight of the airplane, C is the chord of the wings, and 
L is the distance between the center of gravity of the airplane and the 
center of pressure of the horizontal tail surfaces. 

1.5 X2035 X55 

F = = 850 lbs. 

197.5 

234 



Art. 158 Traction and Torque 

850 

or = 26.0 lbs. per sq. ft. for the maximum intensity of 

32.7 

air load on the tail. 

Air Service Specifications for airplanes of this type call for a breaking 
load of 25 lbs. per sq. ft. on the horizontal tail surfaces. This seems 
rather low when compared with the loads as derived by the formulas 
above, but observation has shown that the structures designed to meet 
these loads are at least safe, and so in the analysis a basis of 25 lbs. 
per sq. ft. for the breaking load on the horizontal tail surfaces is used. 
Experimental investigation covering this subject is now under way from 
which it is hoped that considerable information will be derived. This 
load, however, is an ultimate load, while the panel loads just computed 
are working loads. 

Air Service Specifications call for an allowable load at failure cor- 
responding to a load factor of 5. Then the air load on the tail cor- 
responding to the panel loading given by Fig. 158 will be equal to 25 
lbs. per sq. ft. divided by 5, or 5 lbs. per sq. ft. This will increase the 
panel load on one truss at Station 10 bv (5 X 32.7)/2 = 81 lbs., making 
a total panel load of 81 + 29 = 1 10 lbs. 

158. Traction and Torque — The propeller traction is a function of 
the horsepower and the speed of the airplane. In pounds it equals 550 
times the horsepower divided by the velocity in feet per second. In the 
case under consideration the horsepower is 150 and the speed 114 M.P.H. 

150 X 550 X 60 

The traction = — — ■ — 494 lbs. 

114 X 88 
Half of this total traction, or 250 lbs., will be taken by each truss. 
The loading from the propeller torque is found from the formula 

63000 P 

Torque reaction = 

nd 
where P is the horsepower of the engine n the revolutions per minute, 
and d the distance between engine bearers in inches. 

Substituting in the formula the various known quantities we have, 

63000 X 150 

Torque reaction = =455 lbs. 

1600 X 13 

This torque tends to lighten the load on one truss and to increase it 
on the other. The panel loads at Stations 0, 1, and 2 are, therefore, in- 
creased to take care of the torque loads. The revision of these panel 
loads is shown below, and the revised loads for the flying condition are 
shown in Fig. 159. As the values obtained for the torque and traction 
are maximum values, the load factor of 5. which must be applied to the 
static loads dtie to the weight of the airplane, need not be applied to 
them. In order, therefore, to obtain torque and traction loads compar- 
able to the loads shown in Fig. 158 these values are divided by the 

235 



Fuselage Art. 159 

dynamic factor of 5. The torque is distributed in the same proportions 
as the weight of the power unit. 

Station 
From Art. 156 212 

455 379 
Torque X 53 

5 650 

265 lbs. 
Station 1 
From Art. 156 139 

455 197 
Torque X 27 

5 650 

166 lbs. 
Station 2 
From Art. 156 139 

455 74 

Torque X • • 11 

5 650 

150 lbs. 

159. Supporting Forces — The loads shown in Fig. 159 are supported 
at four points by the wing spars. As the distribution of the total load 
between these four points is indeterminate, assumptions must be made. 
In this case it is assumed that 25 per cent of the load is carried by the 
upper wing and 75 per cent by the lower. The lower spar connections 
are considered to carry the majority of the load on account of the lift 
wires being connected to the fuselage at the lower spar hinges. Where 
the wings are internally braced, as in the Fokker D-7, the upper wing is 
assumed to take half or slightly more than half of the load. The load 
carried by each wing is assumed to be divided between the front and rear 
spars in inverse proportion to the distance from the center of gravity of 
the loads to the respective panel points as illustrated by Fig. 160. 

The propeller traction and torque are not at their maximum simul- 
taneously with the maximum tail load, but as the loading on the tail has 
very little effect on those members forward of the cockpit, and, as the 
loading on the forward part of the fuselage has very little effect on the 
members rear of the cockpit, both of these maximum loading conditions 
may be considered simultaneously. 

Sometimes the center of .gravity of the loads falls to the rear of the 
lower rear spar, and in this case the forward and rear portion of the 
fuselage must be figured separately. In computing the rear portion, 
enough load should be added to the forward panel points to bring the 
center of gravity forward of the rear spar. For the forward portion 
enough load should be assumed to be subtracted from the tail load to 
obtain the same result. 

Another method of compensating for the displacement of the center 
of gravity of the loads by the air load on the tail is preferred by some 

236 



Art. 160 



Stresses for Flying Conditions 




Fig. 159. Panel Loads in Maximum Air Load Condition 



C hC^d of tY,n^ \ 



5to ? 




429 1 346 

Fig. 160. Wing Spar Reaction in Maximum Air Load Condition 



designers. This method is to decrease the panel loads to the rear of 
the center of gravity and increase those forward of it so that the center 
of gravity of the air load and the modified panel loads will be at the 
center of gravity of the airplane. In following this method the changes 
in the panel loads should be directly proportional to the distance from 
the center of gravity. This method is theoretically more correct than 
the one followed in this chapter, though the latter is more conservative. 

160. Stresses for Flying Conditions — Fig. 161 shows a line diagram 
of the rear portion of the fuselage with the panel loads indicated, and 
also the stress diagram. 

Fig. 162 shows the stresses in the forward part of the fuselage due to 
the loads shown. It is assumed that half of the traction is taken out at 
each of the two points where the front wing spars are connected to the 
fuselage. Force EF is increased to balance the horizontal component of 
force GH. The cabane wires will exert this force. 

161. Flying Condition with Rudder Turned — The only forces that 
need be considered in this condition are the air load on the fin and rud- 
der and the reaction at the wing spars. The air load may be considered 
as a concentrated load acting on the tail post at the level of the center 
of gravity of the vertical tail surfaces. This is not exact, as the actual 
distribution of load on these surfaces is unknown, but it is a reasonable 



237 



Fuselage 



Art. 161 



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Art. 161 



Flying Conditions with Rudder Turned 



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Fig. 162. Stress Diagram for Front Portion of Fuselage in Maximum 
Air Load Condition 



assumption and a simple one. The tail post will then act as a beam 
loaded at the level of the center of gravity of the vertical surfaces, and 
supported at two points by the longerons. It is then a simple matter 
to compute the horizontal load on each pair of longerons. Very often 
the center of gravity of the vertical tail surfaces will be above the upper 
longerons, in which case these longerons will carry a load greater than 
the horizontal air load, and the load on the lower longerons will be in 
the opposite direction. A stress diagram needs be drawn only for the 
horizontal truss containing the pair of longerons carrying the heavier 
load, as the web members of both trusses will usually be identical. These 
web members are the only ones in the fuselage determined by this condi- 
tion of loading since the longerons are determined by the stresses due 
to the heavier vertical loads. The stresses in the horizontal trusses will 
be less than those computed owing to the action of the diagonal cross 
wires at the panel points which transfer part of the stress from the 
upper truss to the lower. 

239 



Fuselage Art. 162 



In the case under consideration the longerons are 13 in. apart at the 
tail post and the center of gravity of the fin and rudder IS in. above 
the upper longerons. The upper horizontal truss will then carry 
28/13 =2.16 times the horizontal air load. Air Service Specifications 
require an ultimate horizontal tail loading of 25 lbs. per sq. ft. There- 
fore, the upper horizontal truss will have to carry a load at the tail post 
of 2.16 X 25 X 10.1 = 545, say 550 lbs. Fig. 163 shows the line and 
stress diagram of the rear portion of the airplane for this load. It is 
not necessary to carry these stresses further forward as the stresses are 
small and other considerations will govern the design of the members 
in front of those computed. 

162. Landing with Tail Up — The propeller traction and torque are 
negligible quantities, and the load on the tail surfaces is assumed to be 
just enough to keep the line of propeller thrust horizontal. The panel 
loadings for all panels except No. 10 are as shown by Fig. 158. At that 
panel there is an upward tail load balancing the airplane. This can be 
found by taking moments about the point of contact of the landing wheel 
and the ground. 

T.L. X 208 = 865 X 12.5 
865 X 12.5 

Tail load = = 52 lbs. 

208 
Chassis reaction =865 — 52 = 813 lbs. 

The reaction at panels 2 and 4 is found by taking moments about 
the center of gravity. 

0— 52 X208 — R.X27 + R, X21=0 

R 9 + R 4 =813 
R 2 X 21 — 27 X 813 + 27 R, — 52 X 208 = 
48.0 R = 21,951 -f 10,816 = 32,767 
R, = 683 lbs. 
R 4 =130 lbs. 

It would be more exact to subtract the weight of the chassis from 
both the panel loads and the chassis reaction, but this would require 
relocating the center of gravity and would entail unnecessary computa- 
tion with little or no real gain in accuracy. Fig. 164 is the stress diagram 
for this loading. The diagram for the rear portion of the fuselage is 
drawn to a larger scale than the front portion on account of the small 
forces acting. 

In the foregoing the attempt is made to represent dynamic conditions 
by static loads in equilibrium. The vertical reactions figured at the 
chassis strut points do not represent functions of the stresses in the 
chassis struts under the landing condition with the tail up, but are only 
the loads located at the points shown, which will maintain the structure 
in equilibrium. 

The stresses shown in the bottom longerons between the reaction 
points are not, then, as large as they would be under actual conditions 

240 



Art. 162 



Landing with Tail Up 



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242 



Art. 163 Computation of Loads and Stresses for Three Po:nt Landing Condition 

of landing with the tail up. If, however, the connections between the 
fuselage and chassis are made strong enough it is felt that the stresses 
in the bottom longerons will not become troublesome. There are a num- 
ber of strengthening members in the fuselage which are not shown by 
the diagram. These tend to make the problem of distribution of stresses 
in this part of the fuselage indeterminate, but they also strengthen it. 
Hence the conclusion is that to make the connections rigid enough to 
transmit the stresses from the fuselage to the chassis is all that is re- 
quired. 

163. Computation of Loads and Stresses for Three Point Landing 
Condition — When landing with the tail down, the upward reaction on 
the tail skid causes compression in the upper longerons and is the deter- 
mining condition for the design of these members and also for some of 
the web members. 

The most reasonable assumption as to the distribution of the weight 
of the airplane between the chassis and the tail skid is to assume the 
airplane to be a simple beam supported at two points. The weight of 
the airplane should be assumed to act at the projection of the center of 
gravity on the line joining the points of support. No air load on the 
tail needs be considered, and the weight of the wings may be subtracted 
from the total weight of the airplane, as they probably sustain at least 
their own weight for some time after the airplane has reached the 
ground. The tail skid is in contact with the ground at a point behind 
where it is attached to the fuselage, and its load should be represented 
at the point of attachment by a force and a couple. The chassis re- 
actions should be divided between the points of attachment of the chassis 
struts in such a manner that the center of gravity of the upward reactions 
will coincide with the center of gravity of the airplane. Allowance 
should be made for the fact that the chassis reaction is decreased before 
reaching the fuselage by the weight of the landing gear. 

When the tail skid touches the ground in a three-point landing, the 
reaction of the earth can be divided into two components. The vertical 
component resists that part of the weight of the airplane being carried 
by the tail skid. The horizontal component is the friction between the 
skid and the ground. This frictional force tends to make the line of 
action of the resultant tail skid load perpendicular to the line of the pro- 
peller thrust. The force acting on the wheels may also be divided into 
a vertical component and a horizontal component due to friction or in- 
equalities of the ground. But the horizontal component will not be as 
large proportionately as for the tail skid owing to the smaller coefficient 
of friction. These frictional forces, together with the drag, etc., are 
used in slowing down the airplane. Each individual weight in the air- 
plane absorbs a portion of the total decelerating force. The portion 
absorbed by any particular unit may be considered as a small force 
opposing the decelerating force of friction and may be called the "inertia 
load" on the unit. These small inertia loads may be represented by one 
load at the center of gravity. 

Fig. 165 is a diagram showing the loads in this type of landing. N c 

243 



Fuselage 



Art. 163 



and iV ts are the vertical components of the chassis and tail skid reac- 
tions respectively. F c and F ts are the frictional components of these 
reactions. W is the weight of the airplane considered as acting at the 
center of gravity and / is the sum of the inertia loads, also acting at the 
center of gravity. The ratios F ts /N ts , F c /N c , and I/W are un- 
doubtedly different and it would be very difficult if not impossible to 
determine them. They all tend to make the three resultant forces more 
nearly perpendicular to the line of propeller thrust than the vertical 
forces N c , N ts and W are. The assumption sometimes made is that 
these ratios are the same and equal to the tangent of the angle a between 
the ground and the line of propeller thrust. This makes all the loads 
perpendicular to the thrust line. When this is done the tail skid load 
W X L 3 

. If the horizontal loads are neglected, we get the 



is equal to 



tail skid load 




alue. 



Fig. 165. Reactions in Three Point Landing 



The true value is somewhere between the two. As the directions of the 
resultant forces are unknown, and as the assumption that they are per- 
pendicular to the line of propeller thrust gives loads that are on the 
unsafe side, it is recommended that the second assumption, neglecting 
the horizontal components, be used. Since it gives a higher tail skid 
load, it is the safer assumption. This will be done in the present 
example, and all loads will be considered to act at the angle, 90° — a, to 
the line of the propeller thrust. Inspection of Fig. 165 shows that this 
assumption also gives a higher Toad on the front chassis strut, for a 
given value of iV c , than the assumption that the loads are perpendicular 
to the thrust line. In most cases it will give a higher actual value in spite 
of the smaller value of A 7 C . The value of the stress in the rear chassis 
strut will be much too low, but this is unimportant as the maximum 
value of stress in that strut comes in the level landing condition. 

The forces N ts and N c are first computed, and N c is divided between 
the front and rear chassis struts in such a manner as to locate the center 



244 



Art. 164 Summary of Stresses in Fuselage 

of gravity of the supporting forces at the same point as the center of 
gravity of the down loads. The panel loads are assumed to act at the 
line of the propeller thrust. This is allowed for by dividing the panel 
loads between the two longerons. The computation of the supporting 
forces follow. 

The center of gravity of the airplane without wings is 58.5 inches to 
the rear of the propeller hub (see Table XXIV). It may be assumed 
to be on the line of the propeller thrust. This is not exact, but is a rea- 
sonable assumption for an airplane of this type. It is almost impossible 
to compute the exact location of the center of gravity of an airplane from 
the drawings, the only practical procedure being to compute its loca- 
tion as closely as possible and check the computations by experiment 
on the finished airplane. The line joining the points of contact with the 
ground of the landing wheels and the tail skid is 198 inches long (See 
Fig. 156) and the projection of the center of gravity on this line is 22 
inches from the landing wheels. From this we have, 

22 

Tail skid load = 1730 X = 192 lbs. for both trusses 

198 96 lbs. for one truss 

176 

Chassis reaction = 1730 X =1538 lbs. for both trusses 

198 769 lbs. for one truss 
The chassis reaction must be divided into reactions at Stations 2 
and 4. 

R 2 -fR 4 = 769 
Taking moment about R, 

—865 X 23— 96 X 153 + R 2 X 46 = 
19,895 + 14,688 34583 

R..= = =752 lbs. 

46 46 

Taking moments about R., 

865 X 23 — 96 X 199 — R 4 X 46 = 
46 R 4 = 19,895 — 19,104 = 791 
R 4 = 17 lbs. 
Check: 752+ 17 = 769 

Fig. 166 is the stress diagram for the rear portion and Fig. 167 for 
the front portion of the fuselage due to this loading. The panel loads 
are divided between the upper and lower ends of the struts, as shown in 
the diagrams and the weight of the chassis is subtracted from both the 
panel loads and the chassis strut reactions. 

164. Summary of Stresses in Fuselage — Table XXV shows the 
stresses in the fuselage due to the loads assumed. For the nomenclature 
of the members see Fig. 168. The column in Table XXV headed "Sec- 
tion," refers to the types of sections shown in Fig. 169. Where no di- 
mension is shown in Fig. 169, it is indicated in the table. 

It should be noted that the stresses given in the above table for the 
horizontal members 4 — 4, etc., are computed from a load to which a 
load factor of 5 has been applied, while the stresses in the other mem- 

245 



Fuselage 



Art. 164 




246 



Art. 164 



Summary of Stresses in Fuselage 




Fig. 167. Stress Diagram for Front Portion of Fuselage in Three Point Landing 

Condition 

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Fuselage 



Art. 165 





<L±*£*\ 




/6 


*$* r 






c 




1 


1 -J*" 

v /6 
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Fig. 168 (Upper) 



Fig. 169 (Lower) 



bers must be multiplied bv the load factor to give the ultimate stress. 
(See Art. 161.) 

The table shows the material and dimensions of each member as 
shown in the figures. Using the stresses obtained above it is possible to 
compute the factors of safety of the different members. On design 
work the ultimate stresses are figured and the members designed to 
carry those stresses. 

165. Computation of Factors of Safety — The following values have 
been assumed for the properties of the materials used. 



E 

Lbs. per sq in. 



Ultimate Ultimate 

Tensile Compressive 

Strength Strength 



Steel 

Spruce 
Ash 



28.000,000 
1,600,000 
1,600,000 



55,000 
12,000 
16,000 



36,000 
5,500 
7,800 



The ultimate compressive strength in columns is found from the 



formulas P/A = 



Stt 2 E 



2tt*E 



(L/ P ) 



,P J 



where P/A is greater than f/2, and P/A 



where it is less than f/2. The constants in these formulas are 



250 



Art. 165 Computation of Factors of Safety 



those for a column with restrained ends, the degree of restraint being 
represented by the coefficient 2, where the coefficient for a pin-ended 
column would be 1 and a fixed ended column 4. 

Table XXVI shows the ultimate strength of the various members in 
the vertical trusses and also the factors of safetv under the loads figured 



aoove. 






TABL 


E XXVI 






Member 


Ultimate Strength 


Flying Condition 


Three Point Landing 












Land 


ng Tail Up 




T 


c 


Stress 

+325 


F.S. 

39.4 


Stress 


F.S. Stress F.S. 


0—1 


12,800 




1—2 


12,800 




+325 


39.4 






2—3 


17,200 


8,340 


+309 


55.7 






3—4 


17,200 


5,400 


+694 


24.8 


—116 


46.5 


4—5 


17,200 


3,600 


+682 


25.2 


—208 


17.3 


5—6 


13,500 


3,560 


+567 


23.8 


—222 


16.0 


6—7 


12,000 


2,750 


+438 


27.4 


—185 


14.9 


7—8 


10,500 


2,420 


+300 


35.0 


—142 


17.0 


8—9 


9,200 


2,270 


+ 163 


56.5 


— 81 


28.0 


9—10 




2,960 








— 30 98.6 


0—11 




8,300 






—285 


29.2 


11—12 




8,100 






—397 


20.4 


12—13 




8,340 


—691 


12.1 






13—14 




5,400 


—935 


5.8 






14—15 




3,600 


—945 


3.81 






15—16 




3,560 


—683 


5.2 






16—17 




2,750 


—560 


4.91 






17—18 




2,420 


—438 


5.5 






18—19 




2,550 


—300 


8.5 






19—20 




2,700 


—155 


17.4 






2—12 




8,030 








—560 14.3 


3—13 




3,850 


—600 


6.4 






4—14 




10,600 


—322 


33.0 






5—15 




3,980 


—183 


21.8 






6—16 




3,560 


— 86 


41.4 






7—17 




3,140 


— 95 


33.0 






8—18 




2,680 


— 98 


27.3 






9—19 




2,830 


—104 


27.2 






10—20 




3,200 


—110 


29.0 






2—11 


3,400 




+ 155 


22.0 






3—12 


3,800 




+610 


6.2 






4—13 


3,800 




+245 


15.5 






4—15 


3,800 




+325 


11.7 






5—16 


1,900 




+ 153 


12.4 






6—17 


1,900 




+ 146 


13.0 






7—18 


1,900 




+ 161 


11.8 






8—19 


1,900 




+ 165 


11.5 






9—20 


1,900 




+ 183 


10.4 







251 



Fuselage Art. 166 

Table XXVII shows the ultimate strengths of the various web mem- 
bers in the upper horizontal truss and their factors of safety. The stress- 
es shown are ultimate stresses based on a load factor of 5 so the value 
given as the factor of safety is equal to five times the ultimate strength 
divided by the stress due to the loads used. 

TABLE XXVII 

Member Stress Ultimate Allowable F. S. 



4—4 


—567 


3310 lbs. 


29.0 


5—5 


—480 


3470 lbs. 


36.0 


6—6 


—415 


3050 lbs. 


36.5 


7—7 


—360 


2730 lbs. 


38.0 


8—8 


—244 


2620 lbs. 


53.5 


9—9 


—131 


2580 lbs. 


98.5 


4 — 5 


+815 


Veneer Frame 




5—6 


+642 


1900 lbs. 


14.8 


6—7 


+590 


1900 lbs. 


16.1 


7—8 


+ 535 


1900 lbs. 


17.7 


8—9 


+360 


1900 lbs. 


26.4 



166. Discussion of the factors of Safety Computed — The ultimate 
strengths given in the tables in the last article are all computed on the 
gross sections of the members. For tension members, however, the 
ultimate strength should be computed on the net sections. Inspection 
of the factors of safety, however, shows that none of the tension mem- 
bers, except some wires, are critically stressed. In the case of wires, 
however, the ultimate strength was figured on the net area. It is not 
worth while in this case, therefore, to compute the net area of the tension 
members and correct the values of the ultimate strength and the factor 
of safety. 

Air Service Specifications call for an allowable load factor of 5 for 
this type of airplane. Although the formulas proposed to compute the 
probable values of the dynamic loads on an airplane vary considerably, 
experience has shown that an airplane of this type designed for a factor 
of 5 is safe. Inspection of the tables of factors of safety reveals 
two members in which the factor is less than 5. In member 14 — 15 
the factor is 3.81, and in member 16 — 17 it is 4.91. These factors raise 
the question of revision of the size of these members. They do not 
necessarily indicate that the airplane is unsafe, as there are several fea- 
tures of the design which tend to make the strength of the fuselage 
greater than indicated by the figures above. There are a number of 
redundant members which increase the rigidity of the framework. The 
degree of fixity may be greater than that assumed. The degree assumed 
was that corresponding to a value of C = 2 in the column formulas 
(See Art. 24). The air load on the tail will probably never reach the 
assumed value of 25 lbs. per sq. ft., and if it should the entire airplane 
would be moving in a curved path, which would induce inertia loads in 
the fuselage which would decrease the stress in the lower longeron. If 

252 



Art. 167 



Veneer Covered Truss Type Fuselages 



the degree of fixity of the longerons is such that C=2.25 in the member 
16 — 17, that member will show a factor of safety of 5.45. To raise the 
safety factor of member 14 — 15 to 5.0 the degree of fixity of the member 
will have to be raised so that C=2.63. A value of 2.25 for C in member 
16 — 17 is reasonable and no fear needs be felt for the safety of the mem- 
ber. A value of C=2.63 for member 14 — 15, however, is quite high, 2.5 
being considered the maximum value for the longerons in this type of 
construction. It would seem advisable to revise the size of this member. 
If the routing were omitted the member would have an ultimate strength 
of 5160 lbs. and a factor of safety of 5160/970 = 5.46. Although mem- 
ber 14 — 15 appears unsafe from this analysis, the Vought VE-7 has been 
built and flown very successfully, and no trouble has been experienced 
with this member. This may be partially accounted for by the consid- 
erations discussed above, which may have resulted in the member having 
an actual fixity coefficient of 2.63. 

167. Veneer Covered Truss Type Fuselages — A modern develop- 
ment of the truss type of fuselage is the use of plywood instead of cloth 
for covering the fuselage and the use of wood diagonals instead of wires. 
When this is done only one diagonal is placed in each bay in small 
fuselages, though in large airplanes a second diagonal is often needed 
to stiffen the plywood. The principal advantage of this type of con- 
struction from a structural point of view is the greater degree of fixity 
of the compression members. It can be assumed that the joint between 
a member and the plywood prevents the member from slipping along 
the plywood. The longerons, therefore, if they have plywood on two 
sides at right angles to each other, may be considered fixed in both 
directions along their entire length and may be figured to carry the 
ultimate compressive stress for the material, or 5500 lbs. per sq. in. for 
spruce. The members which have plywood on one side only may be 
considered as fixed along their entire length in one direction, and as col- 
umns partially fixed at the ends in the other. The column constant to 
be used in this case will vary between 2 and 3, depending on the details 
of the construction. Owing to the character of the restraint on the mem- 
bers, it is advisable to make the longerons with plywood on both sides 
square, and the other members rectangular with the narrow face in con- 



£ 



c 



X 



y <^- P/i/h/OOd 



Fig. 170 

253 



Fuselage Art. 168 

tact with the plywood. Consideration of Fig. 170 will show why this 
is the case. 

In Fig. 170, A BCD is a strut glued and screwed to the plywood along 
the face AB. The strength of the joint will prevent slipping, so the strut 
will not deflect in the plane XX. The plywood, however, offers little re- 
sistance to deflection in the YY plane owing to its flexibility, and that is 
the plane in which deflection will occur. It is not important, therefore, 
to have a large radius of gyration about the YY axis, though it is neces- 
sary to have it about the XX axis. For a given area the greatest value 
of p about XX is obtained with a rectangle in which AD is greater than 
AB. If this were the only consideration a very thin strip with the edge 
against the plywood would be the ideal shape for the strut, but the dan- 
ger of local failure and the necessity of having glue joints large enough 
to fix the strut in the plane parallel to the plywood, limit the ratio of 
length to breadth of the section. A ratio of about 2 to 1 for small struts 
and about 2.5 to 1 for large ones is satisfactory. 

168. Semi-Monocoque Type — The semi-monocoque fuselages differ 
from fuselages of the truss type in that the loads are not applied to the 
fuselage at panel points, and the loads and their points of application 
may be taken directly from the balance diagram after suitable provision 
has been made for the distribution of the weight of the fuselage, engine, 
and any other loads acting over a considerable length of the fuselage. 
It is best to assume these loads as distributed loads, either uniform or 
uniformly varying, rather than as concentrated loads. 

In semi-monocoque fuselages the bulkheads are assumed to stiffen 
the fuselage so that failure will result more from bending as a whole than 
from local buckling. The solution for a fuselage of this type is indeter- 
minate, but a good approximate analysis can be made. The chief as- 
sumption is that the fuselage will act as a beam. To find the strength 
at any section, therefore, the bending moment should be calculated as 
usual, and also the section modulus of the effective section. In very 
large fuselages where the ratio of plywood thickness to fuselage diameter 
is very small, the effective section consists of the longerons and part of 
the shell near them, the shell being relied on mainly to fix the longerons 
and insure their acting together. In construction with multiple 
longerons, they are placed more closely near the vertical axis of the 
cross section of the fuselage than near the horizontal axis, as the vertical 
loads are more severe than the horizontal. In the smaller fuselages the 
entire shell may be included in the effective area. The plywood layers 
in the shell make different angles with a line parallel to the axis of the 
fuselage and, on this account, their efficiency in carrying stress varies. 
It seems a reasonable assumption that the effective area of each ply 
be assumed equal to the cross sectional area of the ply multiplied by 
the cosine of the angle it makes, with a line parallel to the axis of the 
fuselage. If the plies were parallel and perpendicular to such a line the 
effective area would be the area of the parallel plies. If each ply made 
an angle of 30 degs. with such a line the effective area would be 0.866 
times the area of the plies. The remarks in Art. 167 regarding the shape 

254 



Art. 169 Monocoque Type 

and degree of fixity of members apply also to the longerons in the semi- 
monocoque type of construction. However, owing to the curved section 
of the shell, where multiple longerons are used they may be assumed to 
carry the full compressive stress of 5500 lbs. per sq. in. 

169. Monocoque Type — Where the shell is not braced by longerons 
and where the bulkheads are too widely spaced, failure will not occur as 
a result of simple bending. The tube will buckle on the compression side 
due to the excessive secondary stresses set up by the primary bending 
stresses. The action is similar to that which takes place in a long column 
under compression. There is this difference, however, that in the case 
of the long column the primary stresses are usually caused by the end 
compression and the secondary stresses by the moment resulting from 
the end loads acting with the side deflection of the column as an arm; 
whereas in the case of the monocoque fuselage there usually is little or 
no end compression due to external loads. The initial loadings cause 
deflection of the fuselage and put the fuselage under bending stresses. 
The secondary stresses caused by the moment of these initial stresses 
acting through the deflection cause ultimate failure. 

The calculation of the stresses in a fuselage of the monocoque type 
is an indeterminate problem. The strength of a new fuselage may be 
predicted by comparison with standard fuselages of similar type. The 
best way to analyze this type of structure is to consider it a beam acting 
under bending forces, and to use a low value for the ultimate allowable 
stress, owing to the local buckling. Just what value should be used has 
not yet been determined, but static tests on full size fuselages of this type 
should show what values are correct. In the light of such tests as have 
already been made, a value of 2000 to 2500 lbs. per sq. in. for spruce 
or poplar plywood, and 3000 to 3500 lbs. per sq. in. for birch are recom- 
mended for the ultimate allowable compressive stresses in the plywood 
shells of both monocoque and semi-monocoque fuselages. However, 
where only narrow strips of the shell next the longerons are considered 
part of the effective section, they may be assumed to carry as high a 
stress as the longerons themselves. 



255 



APPENDIX 

170. Prediction of High Speed at Ground — For the purpose of 
making a stress analysis of an airplane it is necessary to determine its 
maximum speed at the ground. The chart shown in Fig. 171 affords an 
accurate and ready means for doing this. It is based on the principle 
that performance is dependent on the pounds per horsepower and pounds 
per square foot of an airplane, modified by the fineness or general clean- 
ness of the design. What is known as the "fineness factor" of an air- 
plane is a number proportional to the cube root of its total L/D at any 
value of K y . From actual official performance tests on a number of air- 
planes their fineness factors were determined. On the basis of the fac- 
tors so obtained, the following ranges are recommended for the various 
general types of airplane. 

Type Fineness Factor 

Single-seater pursuit 105 to 115 

Two-seater fighters and observation airplanes 100 to 110 

Training, small mail or passenger airplanes 100 to 110 

Day bombers, intermediate size mail or passenger air- 
planes with side engine nacelles 93 to 96 

Large bombers or large commercial airplanes with engine 

nacelles 88 to 93 

To fall within the classes indicated the designs can in no case be 
poor. To illustrate this point the actual fineness factors for several 
well-known airplanes are given. 

Curtiss JN-4D-2 97 

Vought VE-7 106 

U.S. D-4 100 

U.S. D-9A 100 

American SE-5 108 

Fokker D-7 103 

Le Pere Triplane 92.5 

Martin Bomber (with bombs) .... 93 

Orenco "D" 108 

Thomas-Morse MB-3 113 

The most important elements in determining the fineness of an air- 
plane are the disposition of the wing surface (whether monoplane, bi- 
plane, or triplane), the number of bays in the wing cellule, the number 
of exposed wires (control cables, empennage bracing wires, etc.), 
whether streamline wires, faired or unfaired cables, number of engine 
nacelles, and, especially, the shape of the fuselage. 

For a more detailed explanation of this method of predicting air- 
plane performance attention is invited to McCook Field Report A.D.M. 
No. 489 (Serial 1221). 

171. Strength Factors for Wings, Fuselage and Tail Surfaces. 

256 



Art. 171 



Strength Factors for Wings, Fuselage and Tail Surfaces 




J$MOC?9C-JOM Jdd CX?/ U/ 6u<pD0 7 P9J.D3JJOQ 



257 



Appendix Art. 171 

TABLE XXV III 

Type Load Factors 



(1) (2) (3) (4) (5) (6) 



I. Single Seater Pursuit for Dav 

Work— Water-Cooled Engine 8.5 5.5 3.5 35 30 7 

II. Single Seater Pursuit for 
Night Work — Air or Water- 
Cooled Engine 7.5 5.0 3.5 30 25 6 

III. Single Seater Pursuit for Dav 

Work— Air-Cooled Engine .'. 8.5 5.5 3.5 35 30 7 

IV. Single Seater Pursuit for 
Ground W T ork — Air or Water- 
Cooled Engine (armored).. *7-3 *4.5-2.5 3.0 30 25 6 

V. Two-Seater Pursuit — Air or 

W T ater-Cooled Engine 7.5 5.0 3.5 30 25 6 

VI. Three Seater for Ground At- 
tack — Air or Water-Cooled 

Engine (armored) 7-3 4.5-2.5 3 .0 25 20 5 

VII. Two Seater Infantry Liaison 
— Air or Water-Cooled En- 
gine (armored) 6-3 4.0-2.5 3 .0 25 20 5 

VIII. Two Seater Night Observa- 
tion — Air or Water-Cooled 

Engine 6.5 4.5 3.0 25 20 5 

IX. Three Seater Army and Coast 
Artillery Observation and 

Surveillance 6.5 4.5 3.0 25 20 5 

X. Two Seater Corps Observa- 
tion 6.5 4.5 3.0 25 20 5 

XI. Day Bombardment 5.5 3.5 2.5 25 20 5 

XII. Night Bombardment, Short 

Distance 4.5 3.0 2.5 20 15 4 

XIII. Night Bombardment, Long 

Distance 4.0 2.5 2.0 15 10 3 

XIV. Training — Air-Cooled Engine 

(radial or rotary) 8.0 5.5 3.5 35 30 7 

XV. Training — Water-Cooled En- 
gine 8.0 5.5 3.5 35 30 7 

In all cases, both computations and static test are based on full 
military load. 
Column 1 Gives the load factor for high incidence condition with center 

of pressure at its most forward position. 
Column 2 Gives load factor for low incidence condition with center of 

pressure at location corresponding to maximum ground 

speed. 

258 



Art. 172 Equations for Continuous Beams 

Column 3 Gives the required negative load factor to be carried by the 

main plane structure in static test, with center of pressure 

at .25 of chord. 
Column 4 Gives the required average load per sq. ft. to be carried by 

the horizontal tail surfaces as determined by the static test. 
Column 5 Gives the required average load per sq. ft. to be carried by 

the vertical tail surfaces as determined by static test. 
Column 6 Gives the required load factor to be carried by the fuselage 

as determined by static test. 
For Types 1, 2, 3, 4, 5, 14 and 15 for the condition of straight dive 
with down load on the front truss and up load on the rear truss a load 
factor of 1^4 is required for biplanes and triplanes in which there is 
adequate incidence bracing or its equivalent, and a factor of 3.0 where 
such bracing is not present as in monoplanes. For the other types of 
airplanes no computations need be made for this diving condition except 
to determine the drag bracing. In computations for this condition, inci- 
dence bracing is to be neglected. 

*Note — Where split factors are given, as, for example, Type IV, 
Columns 1 and 2, the first factor is to be given by the structure as a 
whole; the second factor is to be given by the structure with any one 
structural member removed. 

172. Equations for Continuous Beams — In this section are given 
three-moment equations and deflection formulas for continuous beams 
for all those conditions of loading which are ordinarily used in the stress 
analysis of airplanes. For the derivation of these equations and for- 
mulas reference can be made to the report issued by the Engineering 
Division of the Air Service entitled "Equations for Continuous Beams," 
Serial Number 759. 

Great care must be used in the application of these formulas to give 
the proper algebraic signs to the moments and shears. Art. 2 explains 
the conventions which are followed in this book. 

The nomenclature given below will be used throughout in all the 
formulas. 

m l = the bending moment at support 1. 
m 2 = the bending momen at support 2. 
m 3 = the bending moment at support 3. 
s 2 = the shearing force at a section adjacent to and at the right 

side of the support 2. 
S- 2 = the shearing force at a section adjacent to and at the left 

side of the support 2. 
i 2 = the slope at 2, the angle being measured between the tan- 
gent and the axis to the right of 2. 
i_ 2 = the slope at 2, the angle being measured between the tan- 
gent and the axis to the left of 2. 
s = the shearing force, m = the bending moment, i = the 
slope and v — the deflection at any cross section at a dis- 
tance x from 2. 

259 



Appendix 



Art. 172 



v 1 = the difference in level between the supports 1 and 2. 
v 3 = the difference in level between the supports 3 and 2. 

For ordinary conditions all the supports are assumed to be on the 
same level so that all terms in the formulas involving v 1 and v 3 equal 
zero. When the supports are not on the same level the values of v ± and 
v 3 are positive when the supports 1 and 3 are higher than the support 2, 
and negative when 1 and 3 are lower than 2. 

All these formulas are derived for upward vertical loads which are 
considered as positive. In the case of downward loads it is only neces- 
sary to change the algebraic sign of the loads W x , W 2 , zv 1 or zv 2 , as the 
case may be. 

In the ordinary stress analysis of an airplane structure the only de- 
flections required are those within each span above or below the line 
passing through the supports at the ends of the span. Cases 1A — 5A 
give formulas for such deflections. Their derivation is similar to the 
corresponding cases with two spans. Deflections are positive when up- 
ward and negative when downward. The deflection which is required 
in an analysis is that occurring at the point of maximum moment in the 
span, where the stresses are a maximum. The point of maximum mo- 
ment is the point of zero shear (considering all the loads and moments) 
and can be easily found from the shears. For combinations of loads 
the deflections for the different cases should be added together to ob- 
tain the final deflection. 

The reactions at the supports are found by adding the shears on 
either side of the support, taking account of the algebraic signs given 
for continuous beams. A positive result indicates an upward reaction 
and a negative one the reverse. 

Case 1 — Concentrated Loads. 



4 W, 



4 W 



. 



Je 



G? ,« *' » 



Lz 



f 



LI 



Origin at Support 2. 

W 2 d 9 

m L L 2 +2m 2 (L 1 +L 2 )+m 3 L 1 = (L 2 2 — d 2 2 ) 

L 2 

W.d, 6EIv 3 6EIv t 
+ ■ (Lr-clx 2 )! + 



260 



Art. 172 



Equations for Continuous Beams 



Case 1A — Deflection in One Span for Concentrated Loads. 



1 


- 


w 








. * .1 




d 








c 






L 








1 


1 






7 



Origin at Support 1. 

Deflection for Part of Beam between x = and C. 

m, L s,L 2 W d : 



EL 



IT^X" S X X° 

+ 

2 6 



+ — + - 

6 



Deflection for part of beam between x = C and L. 

m x x 2 s x x a W(x — C) 3 fn^L s^L 2 W d J 

__ i ^ 1 I . 



EIv= 1 1 

2 6 

Case 2 — Uniform Loads. 



6 



I 2 



6L 





t t * t t 








' * I"/ f 






La 


2 Ji 




1 


FT, 


'^ 


'". 



Origin at Support 2. 

w.,L., :; w,L, :; 6EIv, 6EIv 
m 1 L 2 +2m 2 (L 1 +L 2 )-\-m 3 L 2 ==——-\ 1 — -- f 



Li L 2 



261 



Appendix 



Art. 172 



Case 2A — Deflection in One Span for Uniform Loads. 



1 I ♦ * * 



E 



^u 



*ff. 



ffx 



Origin at Support I. 
EIv = x(x — L) 



m 1 Sj_ w 

+ (x+L) +-(x 2 +xL+L 2 ) 

2 6 24 



Case 3 — Uniformly Varying Loads. 




Origin at Support 2. 

7w 1 L 1 3 2w 2 L 2 3 
m 1 L 2 +2m 2 (L 1 +L 2 )+m 3 L 1 = —- - -\ 



6EIv« 



+ 



60 15 

6EIv 3 



262 



Art. 172 



Equations for Continuous Beams 



Case 3A — Deflection in One Span for Uniformly Varying Loads 




Origin at Support 1 



EIv = x(x — L 



r m 1 s x (x+L) w(x+L) (x^L 2 )"! 
— + + 



120L 



m 2 — m 1 w L 



»i- 



L 6 

Case 3B — Deflection in One Span for Uniformly Varying Loads. 



Origin at Support 1. 
EIv=x(x— L) 




m i s i 

- + — (x+L) 

2 6 



f X(5x :l L— x 4 — 4L^) 



120L 



m 2 — m x wL 



Si 



263 



Appendix 



Art. 172 



Case 4 — External Moments Within Spans. 





m fa ma 




M <** bU C * « 


fl C \^ J, k 






L * 


L< 




' 


^t 


Ffz 


f?3 



Origin at Support 2. 

m 1 L 2 -|-2m 2 (L 1 -|-L 2 )-f-m. ; L 1 = m a 



6C X 



3C, 2 



L, 



2L, 



6C. 



3C„ 3 



L. 



2L. 



+ 



6EIv 3 6EIV! 



+ 



The algebraic signs of the moments m a and m b may be deter- 
mined by the following rules: 

(1) Moments in the span to the right of the origin are positive 
when clockwise and negative when counter-clockwise. 

(2) Moments in the span to the left of the origin are positive 
when counter-clockwise and negative when clockwise. 



Case 4A — Deflection in One Span for External Moments Within 
Span. 






iZ-J. 



+ /T* 



264 



Art. 172 



Equations for Continuous Beams 



Origin at Support 1. 

Deflection for Part of Beam between x = and C. 

f m x s t (x+L) "] m a d 2 x 

EIv = x(x — L) I 1 

I 2 6 



2L 



Deflection for Part of Beam between x = C and L. 



m, Sl (x+L)] 



f m x s : (x- 

EIv = x(x— L) | 1 

I 2 6 



9 



d 2 : 



(x-C) 2 - 



Case 5 — External Moments at Supports. 



m< 



"16 



c c 



m t 



% 



*R 



ve* 



3^ 



Origin at Support 2. 

m 1 L 2 +2m 2 (L 1 -]-L.2)+ m :; L, '= — m a L 2 — 2m b L 1 -f 



6EP 



+ 



6EIv x 



L., 



The moments m, and m 2 are positive when clockwise and 
negative when counter-clockwise. 

The derivation of this formula is given below. 
m„ m 2 and m 3 are the moments to the left of supports 1, 2 
and 3 respectively. 

265 



Appendix 



Art. 172 



Case 5A — Deflection in One Span for External Moments at Supports. 



m t 



*r 






^*— 



te. 



+e, 



Origin at Support 1. 

x (x— L) 
EIv = 



6L 



(m^nia) (2L — x)+m 2 (L+x) 



m 1 and m 2 are the moments calculated from Case 5. 

Derivation of Three-Moment Equation for Case 3. 
Consider Span 2 — 3. 



(1) s = s 2 + 



W X X' 



2L, 



( 2 ) m = m 2 + s 2 x + 



s 2 x^ w x x* 

( 3 ) Eli = m 2 x H 1 h k. When x=0, k=EIi 2 . 

2 24L, 

m x 2 s 2 x 3 w^ 5 

(4) EIv = — '■ [-— 1 hEI^x+k,. When x=0, v=0 and 

2 6 120L X 

k x = 0. 

Let x = L x in equation (2) and solve for s 2 , noting that m be- 
comes rrio. 



(5) s 2 =- 



m 3 — m 2 w^L x 



Let x=L x in equation (4) using the value of s 2 from equation 
(5) and solve for EIi 2 . 

266 



Art. 172 Equations for Continuous Beams 



EIv 3 m 2 L 1 m 3 L 1 JwJ-i^ 
(6 ) EIi 2 = + 



L x 3 6 360 

Consider Span 1 — 2. 



( 7 ) m = m 2 +s_ 2 x- 



w x 2 w x J 



6L 2 



S. 2 x 2 w,x 3 w 9 x 4 

( 8 ) Eli = m,xH — | |-k,. When x = 0, k 2 =EIi_ 9 . 

2 6 24L 2 

m 2 x 2 s_ 9 x 3 w 9 x 4 w 2 x 5 

( 9 ) EIv = — — H 1 — hEIi-,x+k,. When x = 0, 

2 6 24 120L 2 

v = and k, = 0. 

Let x = L 2 in equation (7) and solve for s_ 2 . 

m 1 — m, w,L 2 

(10) s. 2 = - — 

L 2 3 

Let x — L 2 in equation (9) using the value for s_ 2 from equation 
(10) and solve for EIL 2 . 

EIv, m,L 2 rrijL., 8w 2 L, 3 

(11) Eli.,- — ■ + - 

L, 3 6 360 

Since the slopes given by equations (6) and (11) are of opposite 
signs, EIi 2 -{-EIi_ 2 =0. 

Adding these equations and transposing 

7W.L, 3 2w,L, 3 6EIv, 6EIv, 

(12) m 1 L 2 +2m 2 X(L 1 +L 2 )+m 3 L 1 = + + + 

60 15 L x L 2 

Derivation of Three-Moment Equation for Case 4. 
Consider Span 2 — 3. 

For Values of x from to C 1 . 

( 1 ) m = m._, -J- s 2 x 

s x 2 
( 2 ) Eli = m 2 xH — 2 —+k When x = 0, k — Eli.,. 

2 

m,x 2 s.,x 3 

( 3 ) EIv = — - — h~ r-EIU+k, When x = 0, v = and k x = 0. 

2 6 

For Values of x from C y to L^ 

267 



Appendix 



Art. 172 



( 4 ) m = m 2 -j-s 2 x-|-m a 

s 2 x 2 

( 5 ) Eli = m 2 x -| \- m a x + k 2 

2 

When x = C 1; equations (2) and (5) are equal because of the 
continuity. 

Equating and solving for k 2 , k 2 = EIi 2 — m a C 1 

-2 



( 6 ) EIv = 



-|-E'.Ii 2 x — m a C 1 x-|-kc 



2 6 2 

When x = C t equations (3) and (6) are equal 

niaC^ 2 
Equating and solving for k 3 , k 3 = 



When x = h 1} EIv = EIv 3 . 

Let x == L x in equation (6) and solve for the value of EIi 2 



(7) Eli,- 



EIv 3 m 2 Lj s^ 2 nial^ 



L, 2 6 2 



+ maCj. 



niaC, 2 



2L X 



Let x = L x in equation (4) and solve for s 2 . 
nio — m, — m fl 



(8) s 2 



Substituting this value of s 2 in equation (7) 



EIv, 



(9) Eli, 



r 2m 2 -(-2m a +mc 



niaCi 2 
L x \- m a C, 



2U 



Treating span 1 — 2 in the same way 



EIv, 



(10) EIL 2 = 



r 2m 2 +2m b -j-m 1 "" 



m b C 2 2 

L 2 h m bQ 

2L„ 



Since the slopes given by equations (9) and (10) are of opposite 
signs EIi 2 -(-EIi. 2 = 

EIv 3 EIv 3 nigLj^ 2m. > (L 1 + L 2 ) n^L, 

(id o = — + — ■ — : 



L, L 2 6 6 

+m a Q 1 + m„ C. 

t 2L, 3 J 

268 



C 2 2 L 2 

2L. 3 






Art. 172 



Equations for Continuous Beams 



(12) m 1 L 2 +2m 2 (L 1 +L 2 )+m 3 L 1 = m, 



3C, 2 
6C X 2L, 



■m, 



3C 9 2 
6C 2 2L 2 



6EIvo 6EI Vl 

+ : + 

Li-. 1^i 



Derivation of Three-Moment Equation for Case 5. 
Consider Span 2 — 3. 

( 1 ) m = m 2 + m b -f- s 2 x 

s. x 2 

(2) EIi= (m 2 + m b )x + - |-k. When x = 0, k = Eli.,. 

2 

X" s X 

( 3 ) EIv = (m, + m b ) h h EIU + k x . When x=0, v=0 

2 6 

and k, =0. 

Let x = Lj in equation (1) and solve for s 2 . 

m 3 — m., — m b 



(4) s 2 



L 



Let x = L t in equation (3) and substitute for s 2 its value from 
equation (4). 

Solving for EIi 2 . 

EIv. m.,L, m b L, m..L, 

(5) EIi 2 = - — 

L, 3 3 6 

Consider Span 1 — 2. 

( 6 ) m = m 2 -f- s. 2 x 

s_.,x'-' 

( 7 ) Eli = m.,xH — h K When x=0, k 2 =EIi_ 2 

2 

m 2 x 2 s. 2 x 3 

( 8 ) EIv — — H h EIL.x+k... When x=0, v=0 and k..=0. 

2 6 



m l — m 2 -f-m a 



(9) s_ 2 = - 



L 2 

Let x = L 2 in equation (8) and substitute for s_, its value from 
equation (9). 

Solving for EIi. 2 : 

269 



Appendix Art. 173 

EIv-l m 2 L 2 m 1 L 2 m a L 2 



(10) EIL 2 = 

L 2 3 6 6 

Since the slopes given by equations (5) and (10) are of opposite 

signs 
EIi 2 +EIL 2 = 

Adding these equations and transposing 

6EIv 8 6EIv x 

(11) m 1 L 2 +2m 2 (L t +L 2 ) +m 3 L 1 = — m a L 2 — 2m b L 1 H 1 

L x L 2 

173. Methods of Determining Deflection of Beams with Varying 
Load and Section — All methods, whether analytical or graphical, depend 
on the proposition that the deflection curve is developed from the loading 
curve by four successive integrations, or, if the moment curve is known, 
from the moment curve by two successive integrations. The principles 
upon which the special methods treated here rest are: the deflection of 
any point x x , measured from a line tangent to the beam at another 
point x , is equal to the moment about x 1 of the portion of the M/I curve 
of the beam between x and x ± , (p. 154, Boyd's Strength of Materials); 
the moment about any point of any number of co-planar forces is 
equal to the product of the intercept on a line through the point and 
parallel to the resultant of the forces, between the strings holding the 
resultant in equilibrium, the pole distance of the force polygon, and a 
factor depending on the scales of the force and funicular polygons, 
(p. 343, Spoffords, Theory of Structures). 

Fig. 172 shows the variation in the moment of inertia of a simply 
supported beam whose deflection curve is desired. It adds but slightly 
to the work if this curve is irregular instead of being a straight line. 
Fig. 173 shows the loading curve for the beam. The force polygon, Fig. 
174, is constructed from the loading curve by dividing the latter into a 
convenient number of parts and considering the area of each part as a 
concentrated vertical force. These different areas are plotted to scale 
on the line 1 — 10 and the various rays drawn to the pole 0, at a perpen- 
dicular distance, h, from the resultant of the forces 1 — 10. In the usual 
manner, the equilibrium or funicular polygon, Fig. 175, is constructed 
from the force polygon of Fig. 174, each string being parallel to its re- 
spective rays in the force polygon. Each ordinate in the funicular poly- 
gon is proportional to the bending moment at that section. To obtain 
the curve of Fig. 176 each ordinate of the moment curve is divided by 
the ordinate of the moment of inertia curve at the corresponding point 
in the beam, and the quotient plotted to a suitable scale. The force poly- 
gon of Fig. 177 is obtained from this modified moment curve in the same 
way that Fig. 174 was obtained from Fig. 173. The area of each section 
is treated as a concentrated load or force applied at the center of the 
section. From the force polygon for the M/I curve is drawn the final 
funicular polygon, Fig. 178. A curve inscribed in this polygon repre- 
sents the elastic curve of the beam and will be referred to as the elastic 
curve. 

270 



Art. 173 Determining Deflection of Beams with Varying Load and Section 



The proof of this construction is as follows: Assume the number of 
sections of the M/I curve used in constructing the force and funicular 
polygons to increase indefinitely. As the area of the individual sections 
approaches zero as a limit, the funicular polygon becomes a curve in- 
scribed in the polygon actually constructed. Each tangent to this curve 
is the string of the funicular polygon between the lines of action 
of the differential areas of the M/I curve each side of the point of tan- 
gency, and is parallel to the rays of the force polygon holding in equi- 
librium the differential area at the point of tangency. Consider any 
tangent to the curve. The intercept on any vertical line between this 
tangent and the elastic curve represents the deflection of the point 
where the vertical cuts the beam, measured from the tangent. This is 
true because the intercept considered is that between the strings of the 
funicular polygon holding in equilibrium the portion of the M/I curve 
between the two points in question. In other words, the ordinates from 
the curve AB to any tangent to it when multiplied by the proper scale 
are ordinates equal to the deflections of the beam measured from the 
tangent in question. Similarly the ordinates from the curve to any line 
are equal to the deflections from that line when multiplied by the vertical 
scale. In short, the smooth curve inscribed in the funicular polygon, 
Fig. 178, is a graph of the elastic curve plotted to inclined co-ordinates, 
and with the vertical scale of ordinates differing from the scale of 
lengths. If the direction of the axis of lengths is found, it is easy to plot 
the position of any line and measure the deflections of the beam from 
it directly from the drawing. 

In the case of a simply supported beam the deflections are desired 
from a line passing through the two supports. To obtain them, draw a 
line through the two points on the elastic curve representing the supports, 
and the ordinates from this line to the curve will give the deflections de- 
sired. This case is illustrated in Fig. 178. In the case of a cantilever the 
deflections are desired from a tangent to the elastic curve at the support, 
and can be found by drawing such a tangent and measuring the deflec- 
tions. This case is shown in Fig. 179. In general, there will be either 
two points at which we know the deflections or one at which we know 
the direction of the tangent, making it possible to draw a line from which 
the desired deflections can be measured. In the case of a wing spar like 
that of the upper wing of the Fokker D-7, the tangent to the elastic curve 
at the center line of the airplane is horizontal and the deflection at the 
cabane struts is zero. The deflections desired are those measured to the 
curve of the funicular polygon from a line intersecting that curve at the 
cabane strut point and parallel to the tangent at the center line. 

In all of this work great care must be used to multiply the ordinates 
measured from the drawing by the correct scale. The calculation of the 
scale is as follows: 

E is in lbs. per sq. in. 

I is in in. 4 

1 in. = q in. for the linear scale. 

1 in. = p lbs. per inch run. 

271 



Appendix 



Art. 173 



10. 



<3.0 




F/o. i i^Moment or /ne/?t//i C(/&i/e 



3 



a 




a.c 



^ r ~T~ I r 

F/emLo#D/NO Cupve 




F/o, 175 Bend /no Moment Equ/l'3/?/um 
Polygon 



272 



Art 173 



Determining Deflection of Beams with Varying Load and Section 




Ho. i uroecc Pol yoo/v 
roe Load/ n e Cuei/r 



Fie. 173, whence 1 in. in 



For Fig. 173, 1 sq. in. = p X q lbs. 

For Fig. 174, let 1 in. = n sq. in. fi 
Fig. 174 = n X p X q lbs. 

If the pole distance for Fig. 174 is h in., the scale for the bending 
moment polygon is 1 in. = n X P X q 2 X h in. lbs. 

If the ordinates of the moment polygon are now divided by the ordi- 
nates to the moment of inertia curve, and the quotients plotted to a scale 
of 1 in. to m in. of modified bending moment ordinates, the scale for 
Fig. 176 is 1 in.=mnpq 2 h lbs. per in. 3 . Therefore, 1 sq. in. for Fig. 176 
= mnpq 3 h lbs. per in. 2 . 

For Fig. 177, let 1 in. = r sq. in. from Fig. 176. Then 1 in. in Fig. 
177 = rmnpq 3 h lbs. per in. 1 '. If the pole distance for Fig. 177 is h x in. 
the scale of ordinates for the deflection polygon is 1 in. = rmnpq 4 hh 1 lbs. 
per in. The value of the ordinates multiplied by this scale must be 
divided by the modulus of elasticity of the material in the beam to get 
the true deflection. A dimensional equation shows that the above scale 
is correct. 



273 



Appendix 



Art. 173 





I 
Si 



274 



Art 173 



Determining Deflection of Beams with Varying Load and Section 





Fig. 179. Deflection of a Cantilever Beam 



Since it is usually more convenient to compute the values of M/l 
than to obtain them graphically, the M/l curves may be plotted from 
calculated values. The scale, s, used in plotting this curve equals the 
vertical scale for Fig. 176 1 in. = mnpq 2 h = s. The remainder of the 
work is the same as that shown in Fig. 178, but it should be remembered 
that the value of m?ipq 2 h is to be replaced in subsequent formulas by the 
value of the scale, s, for M/L The scale of ordinates for the deflection 
polygon becomes 1 in. = rsq~h x . 

The graphical method may be checked roughly by using the average 
moment of inertia of the beam and computing the deflection by the ordi- 
nal}' deflection formulas. 

Any difficulty there may seem to be in determining the correct scale 
factor is removed if it is understood that in all the scaling and plotting 
the ordinates and areas dealt with are true lengths and actual areas in 
inches and square inches, respectively. As each diagram is drawn the 
scale that was used for it is noted, and not till the deflection is obtained 
is it necessary to compute the scale, which is, therefore, done in a single 
operation. 

If the example given for a singly supported beam had been solved 
analytically, using an average load and moment of inertia, the resulting 
deflection is in error by somewhat less than 3% on the unsafe side. 



5 wk 



5 X $ X 100 4 1,040,000 



384 EI 384 X 6.25 E 



E 



In the case of a beam with the same loading and length but with the 
moment of inertia varying from 10 to 1 the deflection by the analytical 
method, with averaged loads and moments of inertia was again too 
small bv about 8%. 



275 



Appendix 



Art. 173 



In airplanes such as the Fokker D-7, the upper and lower wings are 
internally braced cantilevers. The deflections of the spars are equal- 
ized by the use of an "N" outboard strut. In the Fokker D-7 the spars 
are not the same size, and hence, without the outboard strut would not 
deflect equally. Therefore, as a strut is used, some of the spars take 
part of the load from the others. Before the final design can be made 
of the spars the distribution of load between the spars must be deter- 
mined. This can be done as follows: 

1. Determine by the graphical method the deflection of points c and 
d, Fig. 180, due to the distributed load and the reaction at d, and the 
deflections at a and b due to the distributed load and the reaction at a. 
These deflections should be measured from tangents to the elastic curves 
at e and / as these are both horizontal by symmetry. The deflections 
are measured from these tangents because they are the only ones to the 
two curves which are known to be horizontal. The points e and /, how- 
ever, are not fixed vertically, so the deflections at the fixed points d and 
a must be subtracted from those at c and b to find the absolute deflec- 
tions of the latter. This may be done graphically as described above by 
measuring the deflections from lines through d and a and parallel to the 
tangents at c and /. When the center section of the lower spar is en- 
closed in a sleeve allowance must be made in the M/I curve for the in- 
creased moment of inertia of the spar unit. 

2. Subtract the absolute deflection of b from that of c to find the 
relative movement of the two points. 

3. Apply a concentrated load of unity to each beam at the strut 
point, and determine the deflection of each beam at this point due to the 
unit load. 

4. The relative deflection of the beams of either truss at the strut 
point produced by the distributed load, divided by the numerical sum 
of their deflections due to the load of unity equals the value of the stress 
in the strut, or the amount of support that one beam affords the other. 

5. Find the relative deflection of the front and rear trusses and the 
forces necessary to equalize them by the same method as above. 



c 




Fig. 180 

276 



Art 173 Determining Deflection of Beams with Varying Load and Section 

6. Figure the resultant forces acting on each spar at the N strut and 
correct the total bending moment in the spars by the moments due to 
the forces. 

There are several designs that have been built or proposed in which 
the wings are unbraced cantilevers. As far as strength is concerned, 
such wings can be readily designed, but in the opinion of many engineers 
this type of construction is limited by the large deflections of the wing 
tips. Whether or not this deflection is detrimental to the flying qual- 
ities of the airplane, the determination of the deflection curve of the wing 
spars is an essential part of the stress calculation. 

In addition to the graphical solution, the deflections may be ob- 
tained analytically. Essentially the graphical method consisted in the 
determination of the moment curve of a beam under a loading repre- 
sented by the M/I or M/EI curve. This can be done by any of the 
analytical methods of finding bending moment from the load curve. 
Care must be taken, however, to make sure that the deflections obtained 
are those desired. In the case of a cantilever the moment desired is that 
of the area of the curve between the given point and the point where the 
axis of the beam is horizontal, about point x. In the case of a simply 
supported beam the moment of the area of the M/EI curve between a 
point and a support will give the deflection from the tangent at that sup- 
port. To find the deflection from a line passing through the supports it 
is necessary to know the slope of this tangent which can be found by 
dividing the deflection of the other support from it by the span. Know- 
ing these values it is a simple matter to find the deflection desired. 

If the work is done analytically it is best to keep the work in tabular 
form as far as possible and to find the moments by the formula in Art. 
19. By using this formula it is a simple, though somewhat tedious, mat- 
ter to find the moment curve of any M/I curve no matter how irregular. 
The M/I curve should be divided into a convenient number of parts, 
each one of which may be considered as a trapezoid or even a rectangle 
with very little loss of precision, and the moments computed at each 
division point. In order to illustrate the method of using the formula of 
Art. 19, the moments of the example in Art. 17 are worked out below. 



1 


2 


3 


4 


5 


6 


7 


8 


Sta. 


S 


X 


Sx 


F 


a 


Fa 


M 


A 


—100 



















25 


—2500 


—150 


12.5 


—1875 




B 


387.6 












—4375 






20 


7752 


—120 


10.0 


—1200 




C 


217.6 












2177 






25 


5440 


—150 


12.5 


—1875 




D 


22.4 












5742 






40 


— 896 


— 40 


20.0 


— 800 




E 


—262.4 












4086 






15 


—3936 


— 15 


7.5 


— 112.5 




F 















— 2.5 



277 



Appendix Art. 174 



Col. 2 gives the shear to the right of the station in col. 1. 

Col. 3 gives the distance between consecutive stations. 

Col. 4 gives the product of the distance in col. 3 on the same line and 
the shear in col. 2 in the next line above. 

Col. 5 gives the amount of load in the distance represented by the 
figure in col. 3. 

Col. 6 gives the distance from the center of gravity of this load to the 
station at which the moment is being computed. 

Col. 7 is the product of cols. 5 and 6. 

Col. 8 gives the moment at all the stations on the same line in col. 1. 
It is the algebraic sum of the value just above in cols. 8, 4 and 7, and 
is obtained from the formula of Art. 19: Mx = M b -f- S b x + F «a. 

These values do not agree exactly with those in Art. 17 because the 
values of the reactions are found to one more significant figure. The 
value of the moment at F should equal zero. That it does not is due to 
the fact that the reactions were figured only to the nearest tenth of a 
pound. If they had been figured only to the nearest pound the error 
would have been 37.5 in. lbs. instead of — 2.5 in. lbs. 

In the case of figuring deflections, there are no concentrated loads or 
reactions and the value of the shear at any point is the sum of the ele- 
mentary areas up to that point, i.e., S = %F. The work can easily be 
arranged in similar tabular form. For a cantilever this method is prob- 
ably as easy as the graphical method, though for simply supported beams 
the advantage may be with the latter. The point to be stressed is that 
both methods are basically the same and the work may be carried on, if 
desired, by almost any combination of the two methods, some steps 
being taken graphically and some analytically. Up to and including 
the computation of the values of M/I or M/EI, the analytical method 
is almost always preferable. 

The terms M/I curve and M/EI curve have been used interchange- 
ably in the above discussion. This is allowable if it be assumed, as it 
nearly always is, that E is a constant. In that case exactly the same 
deflections will be obtained by using the M/I curve and dividing its 
moment by E as by using the M/EI curve. Two advantages of the 
former method are that the numbers used are generally greater than 
unity, instead of small decimals, making them more convenient to figure 
with, and that fewer divisions by E are usually necessary, as in nearly all 
cases the ordinates of more points on the M/I curve must be determined 
than deflections. 

174. Properties of Woods at 10 Per Cent Moisture — The properties 
of wood, which have been in common use, have been based on a moisture 
content of 15 per cent. As most specimens of woods in actual use on land 
airplanes average slightly less than 10 per cent moisture content, the 
strength properties given in Table XXIX were reduced to this value. 
For seaplanes and flying boats, however, all strength properties should 
be based on 15 per cent moisture. The woods included in this table 
were taken from Signal Corps Specification 15020-B. Only those species 
listed in Column 2 of Table XXIX are permitted by the above specifi- 

278 






Art. 174 Properties of Woods at 10 Per Cent Moisture 

cation. The strength properties are based on the values given in Bulletin 
No. 556 of the Forest Service of the U. S. Department of Agriculture 
which, as far as data were available, are adjusted to 10 per cent moisture 
content by the formula s 2 = j 1 »10 a (mi—m 2 ) in which s t is the strength 
value as found from tests at m 1 per cent moisture, and s 2 equals the 
strength value adjusted to 10 per cent moisture; a is a constant derived 
from experiment. This formula is the result of work done at the Forest 
Products Laboratory. In the cases where values for a could not be ob- 
tained the method of adjustment given in Bulletin No. 556 was followed. 

With the exception of elm all the values in Table XXIX are averages 
of the properties of each of the species, listed in Column 2, of the wood in 
question. In the case of elm the properties of rock elm only were con- 
sidered. As the average specific gravities of slippery and white elm are 
.54 and .51, respectively, while the specific gravity requirement in 
Column 3 for elm is .60, any specimens of slippery or white elm meeting 
this requirement will have strength values equal to those of rock elm. 
It should be further noted that the values for the moduli of rupture in 
Column 7 are for rectangular sections only. With routed sections a 
large decrease in the modulus of rupture occurs, which varies with the 
degree of routing. Tests have been conducted to determine quanti- 
tatively the value of this decrease. Three series were run, the first with 
very slight routing, the second with normal routing, and the third with 
excessive routing. There were ten routed specimens in each series 
except series two, in which there were fifty, and for each routed specimen 
there was a matched solid specimen. All specimens were of the same 
outside dimensions, 2^4 x V/2 in. As a result of a study of the data 
from these tests, the following empirical formula was obtained, which 
is believed to give reasonably accurate values for the modulus of rup- 
ture of spruce or fir spars of a normally routed I section: 

1.12 F S I S 
F r = - 



Ir 

F,. = modulus of rupture of routed section. 

F s = modulus of rupture of solid section. 

I,. = moment of inertia of routed section. 

I s = moment of inertia of a solid rectangular section 
of same area and center height as the routed sec- 
tion. 

Further, experimental work is being done to verify this formula, and 
also to determine the effect of routing on strength in horizontal shear, 
which is reduced probably even more than the modulus of rupture. The 
value of 750 lbs. per sq. in. that is given below for spruce is for routed 
sections. In calculating the shear in solid sections, such as is produced 
by bolts which are subjected to pull parallel to the grain of the wood, as 
in a longeron splice, the values of the shear strength listed in Col. 12, 
Table XXIX may be used. 

279 



Appendix Art. 174 



For the reasons that spruce is so largely employed for highly stressed 
members in airplane structures, and that each of the three species of 
spruce are in common use, the working values for the various properties 
recommended for design work are not the average of these properties 
for the three species, but, in the case of the ultimate compressive strength 
parallel to the grain, are the lowest of the three values, and in the case 
of the modulus of elasticity and ultimate horizontal shear are values 
which experience has shown to be suitable. 

STRENGTH VALUES FOR SPRUCE 

Modulus of Modulus of Ultimate Compressive Horizontal 

Rupture Elasticity Strength Parallel to Grain Shear 

Lbs. per sq. in. 

~la300~~ 1,600,000 " 5,500 750 

In Fig. 181 is given a curve showing the reduction in stress that is 
necessary to allow for column action in spruce columns. Similar curves 
for other woods may be constructed by the aid of the formulas in Fig. 
181. Interplane struts as a rule fall in the class of slender columns, and 
their strength may, therefore, be computed by Euler's formula for pin- 
ended columns. This formula is not applicable to spruce or fir struts 
with an L/p less than 80 to 90. For center section struts use Johnson's 
formula or the curve in Fig. 181 for pin-ended columns. 

Where members, such as wing beams, are subjected to both bending 
and compression, the following method of calculating ultimate allowable 
stresses is recommended. 

f„ 

(F-C) 



(, 



+ C 



.fb + fc 

f a = ultimate allowable stress 

f b = calculated stress due to bending. 

f c = calculated stress due to compression. 

F = modulus of rupture. 

C = maximum allowable P/A as determined from 
the curve for pin-ended columns in Fig. 181. 
When considering a point near mid-span the length of the column 
may be conservatively taken as equal to the span. At inner strut points 
where column action is present the column length may be taken as equal 
to .25 of the span on one side of the strut plus .25 of the span on the 
other. At cantilever strut points C equals the full compressive strength 
of the material. 

The weights in Col. 4 of Table XXIX are calculated by using the 
specific gravities, given in Bulletin No. 556, which are averages for the 
species and are based on oven dry weight and air dry volume, to com- 
pute the weight of a cubic foot of dry wood. This weight is then in- 
creased by 10 per cent to allow for the moisture. 

280 



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281 



Appendix 



Art. 174 



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282 



m« 


fll 


H + 

[20j 


15^ 






Art. 174 Properties of Woods at 10 Per Cent Moisture 



THE EFFECT ON THE STRENGTH PROPERTIES OF WOOD 
OF SPIRAL AND DIAGONAL GRAIN 

In determining the true slope of grain in any piece of wood two kinds 
of sloping grain must be considered, namely: (1) diagonal grain or the 
slope of the annual growth layers with respect to the axis of the piece 
and (2) spiral grain or the slope of the wood fibers within these annual 
layers. Diagonal grain is caused by the fact that the stock is not sawn 
parallel to the annual growth rings. Its maximum value occurs on a 
radial face. Spiral grain is a maximum on a face tangential to the 
growth rings. Frequently in inspection when a piece has both spiral and 
diagonal grain it is wrongly classified as having a grain with a slope 
equal to the slope of either the diagonal or spiral grain, depending on 
which has the greater slope. The point should be emphasized that when 
two slopes are present the resultant slope is greater than either single 
slope. Classification on the basis of absolute or resultant slope is the 
only safe and correct method. Suppose the slope of the diagonal grain 
is 1 in 15 and that of the spiral grain 1 in 20, the absolute slope = 

1 

— or a slope of 1 in 12. Classification bv the 

12 

method of greater slope may admit material 15 per cent lower in average 
strength properties. 

Tests show that the damaging effect of combined spiral and diagonal 
grain is the same as that of spiral or diagonal grain of the same absolute 
slope, and that spiral grain is no more harmful than diagonal grain of 
the same degree. However, it is more likely to escape detection, and 
spiral grained material is more subject to deterioration from season 
checks than diagonal grained. 

Table XXXI gives quite complete data on the effect of spiral grain 
on the strength properties of Sitka spruce and Douglas fir. Plots show- 
ing the decrease in the value of the different properties bring out that 
some properties are affected by slightly sloping (1 in 35) grain, that 
others are not influenced appreciably until the slope becomes 1 in 15 or 
greater. In general, the decrease is gradual until a slope of from 1 in 20 
to 1 in 15 is reached. In this zone there is a sharp change in the slope 
of the curves and the rate of decrease becomes much more rapid. Slop- 
ing grain affects the modulus of elasticity least and the work and impact 
properties most. Hence, in material for slender struts sloping grain is 
not nearly as important as in members subject to severe shock and vi- 
bration. For most types of airplane construction it may be stated that 
when the absolute slope exceeds 1 in 30 the design must be changed to 
allow for the decrease in the ultimate stresses, and in no case should 
material be used in which the absolute slope is greater than 1 in 15. 
This ruling applies to highly stressed members. 

It is suggested that spiral or diagonal grained stock can be effectively 
used in the webs of built-up spars, or as the center lamination of a strut 
built up with three laminations. 

283 



2 


.0 


1 


.5 


2 





2 





1 


5 


1 






Appendix Art. 174 



Few data are available on the effect of sloping grain on the strength 
properties of woods other than spruce or Douglas fir, and until such data 
are available, the percentage of reduction obtained by tests on spruce 
and Douglas fir should be used. Recent tests show that ash is affected 
by sloping grain to the same degree as spruce and Douglas fir. 

THE RELATIVE MERITS OF WOODS 

In Table XXX the woods are arranged in their order of relative 
merit based on their properties per unit weight. In calculating this 
table certain properties were weighted as follows: 

Work to maximum load in static bending 

Fiber stress at elastic limit in static bending. . 

Modulus of rupture 

Modulus of elasticity 

Fiber stress at elastic limit in impact bending. . 

Ultimate compressive strength parallel to grain 

10.0 
Work to maximum load represents the ability of the timber to absorb 
shock, after the elastic limit is passed, with a slight permanent or semi- 
permanent deformation and with some injury to the timber. This 
property is a measure of the combined strength and toughness of a 
material under bending stresses, and is, therefore, of great importance 
in airplane parts which are subjected to severe shock as in chassis struts, 
to large, suddenly applied loads, or to severe vibration. On small high- 
powered machines toughness is particularly essential. A wood may be 
high in its elastic properties but if it is low in toughness it will not be 
as suitable for most airplane construction as a somewhat weaker, but 
tougher wood. A wood lacking in toughness is said to be brashy. 
Brashness may be either natural or artificial. Wood may be said to be 
naturally brash, when one of its characteristic features is low work 
properties. However, any species may contain brash material due to 
obtaining lumber from very old mature trees, or to its being affected 
by fungus growth or decay. Wood may be said to be brash due to 
artificial causes when it has been subjected to high temperatures or very 
rapid drying. 

The stresses that are developed in service should always be within 
the elastic limit of the material. Occasionally, however, under very se- 
vere conditions the elastic limit will be exceeded, and if the load con- 
tinues for more than a short time gradual failure will ensue. Since 
working stresses are ordinarily based on ultimate stresses rather than 
on elastic limit values, a material with an elastic limit close to its ulti- 
mate strength is much superior to a material of the same ultimate 
strength but with a low elastic limit. 

The weights given the different properties in determining the rel- 
ative merit of the woods, as shown in Table XXX, were chosen with 
the purpose of selecting woods best fitted for general airplane work. 
In wing beams, all the properties which were considered in preparing 

284 



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13 «.- 



Appendix Art. 174 

TABLE XXXI 

RATIOS OF STRENGTH VALUES OF SPIRAL AND DIAGONAL 
TO STRAIGHT GRAINED MATERIAL 

Impact 
Static Bending Bending 



Slope of Fiber Stress Modulus Modulus Work to Maximum 

Grain at Elastic of of Maximum Drop* 

Limit* Rupture* Elasticity* Load* 



SITKA SPRUCE 

Absolute Slope Classification 

1:40 or less 100.0 88.2 100.0 89.4 100.0 92.3 100.0 75.0 100.0 76.2 



1:25 


97.8 


87.4 


97.4 


882 


97.6 


90 5 


86.0 


65.6 


92.1 


70.6 


1:20 


97.0 


87.0 


96.2 


85.9 


96.1 


88.6 


78.9 


57.8 


87.3 


68.2 


1.15 


96.3 


86.7 


92.1 


81.2 


93.0 


85.2 


67.2 


46.8 


77.8 


57.1 


1:12.5 


94.4 


84.8 


88.7 


77.5 


90.7 


82.6 


56.2 


39.8 


67.4 


48.4 


1:10 


89.6 


79.6 


83.0 


72.0 


86.8 


78.5 


45.3 


32.0 


55.5 


39.7 


1:7.5 


81.8 


70.7 


72.4 


60.8 


78.8 


70.8 


34.4 


24.2 


43.6 


30.9 


1:5 


66.2 


55.9 


55.8 


45.4 


64.4 


55.2 


24.2 


15.6 


30.9 


22.2 








DOUGLAS FIR 
















Absol 


ute Slope CI 


assification 










1:40 or 1. 


ess 100.0 


88.3 


100.0 


89.0 


100.0 


87.9 


100.0 


80.2 


100.0 


87.6 


1:25 


96.4 


83.7 


93.0 


83.6 


96.2 


85.7 


83.4 


66.2 


99.0 


84.1 


1:20 


93.2 


80.4 


89.6 


78.4 


94.2 


83.9 


76.1 


57.9 


95.5 


81.1 


1:15 


88.1 


74.7 


84.8 


70.8 


91.5 


80.9 


66.2 


48.0 


87.1 


72.6 


1 :12.5 


83.9 


70.7 


81.8 


65.9 


89.6 


78.1 


60.4 


42.2 


79.6 


64.7 


1:10 


77.9 


65.3 


75.4 


59.2 


86.4 


74.2 


53.8 


35.6 


69.2 


53.2 


1:7.5 


69.2 


56.6 


64.6 


50.6 


79.4 


66.2 


44.7 


28.9 


51.8 


38.3 


1:5 


54.3 


44.6 


46.0 


41.2 


59.6 


49.0 


32.3 


21.5 


35.3 


21.9 



*The values in the first column under each property are the ratios of the average 
of the spiral or diagonal grained material to the average for straight grained 
material, expressed in per cent. The values in the second column under each 
property are the ratios of the "most probable value below mean" of the spiral 
or diagonal grained material to the average for straight grained material ex- 
pressed in per cent. "The most probable value below the mean", corresponds 
closely with the arithmetical average of all the specimens having properties 
lower than the average properties of all the specimens. It is an indication of 
the reliability or variability of spiral and diagonal grained stock. 

**A11 wood having a slope of grain of 1 to 40 or less was considered as straight 
grained. 

Table XXX are of importance, and, therefore, the order of merit as in- 
dicated by this table is closely correct for this type of member. But in 
other portions of the structure only certain properties are of importance. 
For example, in selecting the best wood for a tail skid only the modulus 
of rupture, the work to maximum load and the fiber stress at the elastic 
limit in impact bending need be considered. In the case of a slender 
strut the modulus of elasticity is the most important property, though 
consideration should also be given the work to maximum load. For 

286 



Art. 174 Properties of Woods at 10 Per Cent Moisture 

short struts, on the other hand, the ultimate compressive strength be- 
comes more influential than the modulus of elasticity. It should be 
noted that in certain members, especially slender struts, where the 
strength varies as the moment of inertia, that in comparing woods of 
different specific gravity it is incorrect to use the first power of the ratio 
of the weights of the woods. The heavier species gives a smaller strut 
for the same strength, but, since the strength varies directly as the cube 
of the diameter, the material in the smaller strut is not so effectively 
placed, thus decreasing the advantage of the heavier, stronger material. 
In designing members subjected to different kinds of stresses an engineer 
should decide what properties are important and what their relative 
importance is. By the aid of the comparative values in Table XXX 
he can then quickly determine which wood is best adapted to any par- 
ticular purpose, always bearing in mind that on large, heavy airplanes 
the shock absorbing properties are less essential than on fast, high- 
powered pursuit airplanes where vibration is severe. Column 10 of this 
table shows numerically the relative advantages of the different woods 
as far as their strength properties alone are concerned. Other factors 
besides strength properties must be considered also. 

COMMENTS ON DIFFERENT SPECIES 

Characteristics other than strength must be considered in deciding 
on the value of a wood for a certain purpose. The comments which 
follow are based on a report by Mr. J. A. Newlin, in charge of timber 
mechanics at the Forest Products Laboratory. 

A material should dry readily, stay to its place well, and not develop 
serious defects such as shakes, checks or rot. The way a wood works 
under the tool, the finish it takes, and all manufacturing conditions 
must be considered, as well as the size of the trees and the defects nor- 
mally found in the species. Frequently high grade stock of an inferior 
species is better than low grade material of a superior species. 

Port Orford Cedar — Recent data on this species is not quite as favor- 
able as that given in Table XXIX. 

Douglas Fir — This wood is considerably harder to dry than spruce 
and is more inclined to shakes, and to check during manufacture, and to 
develop these defects in service. It is inclined to break in long splinters 
and to shatter when hit (a serious defect in certain types of military air- 
planes). The use of Douglas fir in the manufacture of wing beams 
will require considerably more care than is necessary with spruce, but it 
should give very excellent results when substituted for spruce in the 
same sizes. 

Western White Pine — It is more difficult to dry than eastern white 
pine, but could be very satisfactorily substituted for spruce in spruce 
sizes. 

Eastern While Pine — Tests show this wood to be somewhat below 
spruce in hardness and rather low in shock resisting ability. It, how- 
ever, runs quite uniform in its strength properties, is very easily kiln 

287 



Appendix Art. 174 

dried without damage, works well, stays to its place well and is recom- 
mended as a substitute for spruce in spruce sizes. 

Black and Commercial White Ash — These woods are well adapted 
to steam bending. 

Basszvood — Basswood is one of the best species to receive nails with- 
out splitting and is used extensively for webs, veneer cores, etc. 

Birch — Sweet and yellow birches are quite heavy, hard and stiff. 
They have a very uniform texture and take a fine finish. On account 
of their hardness and resistance to wear they can be used to face other 
woods to protect them against abrasion. Birch is extensively used in 
propeller and plywood construction. 

Black Cherry — This is a very desirable propeller wood. 

Poplar — Of the three species listed under this name, cucumber is con- 
siderably the best, while yellow poplar is the weakest. Cucumber is one 
of the few hardwoods which gives promise of being a good substitute for 
spruce. Magnolia is similar to cucumber. Yellow poplar, while rather 
low in shock resisting ability, has good working qualities, ability to retain 
its shape, and freedom from checks, shakes and such defects, char- 
acteristics which make it a fairly satisfactory substitute for spruce. It 
presents no manufacturing difficulties. Poplar is used to some extent in 
propeller construction. 

Elm — This species is low in stiffness though very resistant to shock. 
It steam bends well, and if properly dried can be used for longerons as 
a substitute for ash which is somewhat lighter. Considerably more care 
is necessary in the drying of elm in order to have it remain in shape, as 
it twists and warps badly when not held quite firm. White and slippery 
elm when of the same density as rock elm may be substituted for the 
latter. A lighter grade of white elm could probably be used to excellent 
advantage in the bent work at the end of wings, rudders, elevators, etc. 

Sugar Maple — Sugar maple is quite heavy, hard and stiff. It should 
be used along with birch in propeller manufacture. It has very uniform 
texture and takes a fine finish. Like birch it is often used to protect 
other woods against abrasion. 

Oak — The oaks need not be considered as substitutes for spruce, but 
they play an important part in the manufacture of propellers. They are 
all quite heavy and hard, and are extremely variable in their strength 
properties. White oaks, as a rule, shrink and swell more slowly with 
changes in the weather than do red oaks. Because radial shrinkage in 
oaks is about half the tangential shrinkage, quarter-sawn oak is much 
superior to plain sawn oak for propeller construction. Southern grown 
oaks are much more difficult to dry than are the northern oaks. North- 
ern white oaks when quarter sawn and carefully dried give very satis- 
factory propellers. Quarter sawn northern red oaks are fairly satis- 
factory for propeller construction, but they have the disadvantage of 
being more subject to defects in the living tree, decay more readily, and 
change more rapidly with change in weather conditions. - 

288 



Art. 175 Properties of Duralumin 



Black Walnut — This wood need not be considered at all as a sub- 
stitute for spruce. But it probably makes the best propeller of any of 
the native species. Black walnut is somewhat difficult to dry, but has 
very excellent power of retaining its place and has good hardness to 
resist wear. 

Summary — Data available indicate strongly that the following species 
can be substituted for spruce in highly stressed parts: Port Orford 
Cedar, Coast Type Douglas Fir, Amabilis, Grand, Noble and White 
Firs, Eastern and Western White Pine, Yellow Poplar, Cucumber and 
Magnolia. Certain other woods give good promise of furnishing spruce 
substitutes, though more work is necessary to overcome known difficul- 
ties before they can be definitely recommended; viz, Lodgepole Pine, 
Norway Pine, and Redwood. 

175. Properties of Duralumin — The chemical composition of the 
duralumin made by the Aluminum Co. of America, is as follows: 

Aluminum 94.5 per cent 

Copper 4.0 per cent 

Magnesium 0.5 per cent 

Manganese . 0.5 per cent 

Impurities about 0.5 per cent 

These impurities must contain no lead, tin or zinc. 

The important physical properties are: 

Specific gravity 2.8 

Melting "point" 650° C. 

Ultimate tensile strength 55,000 lb. per sq. in. 

Yield point in compression 27,000 lb. per sq. in. 

Ultimate shearing strength (rivets) 35,000 lb. per sq. in. 

Ultimate bearing strength (rivets) 90,000 lb. per sq. in. 

Modulus of elasticity 10,000,000 lb. per sq. in. 

Elongation in 2 in 18 — 22 per cent. 

Duralumin must not be bent around a pin with a radius less than 
four times the thickness of the duralumin. 

To prevent cracking all edges should be rounded and, when possible, 
sections should be so designed that the edges will be at or near the 
neutral axis. 

Bent plate fittings, with bent lugs which must resist vibration 
should be made from sheet steel instead of duralumin. For stressed 
parts, which while in flight are exposed to an increase in temperature 
of more than 100 deg. C, the use of duralumin is objectionable unless a 
correspondingly smaller strength value is used in computations. Cold 
has no harmful influence on duralumin. The joint between iron and 
steel and duralumin can be made without electrolytic action occurring. 
Pieces, which for better working must be heated, must in all cases be 
re-tempered after completion. 

289 






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!91 



Appendix 



Art. 176 



176. Miscellaneous Tables and Charts. 




Standard Lugs 



TABLE XXXIII 
PROPERTIES OF STANDARD LUGS 



Tie Rod 




Sh 


ank 




Eye Pin 


Rating 




w 


Ultimate 


R 





Ultimate 


D 




3/32 


5/16 


Strength* 

1,610 


7/32 


3/64 


Strength* 

1,205 




1,000 


3/16 


1,900 


1/8 


11/32 


2,360 


9/32 


1/16 


2,470 


3/16 


2,600 


1/8 


15/32 


3,220 


3/8 


5/64 


3,330 


1/4 


3,400 


3/16 


13/32 


4,180 


11/32 


1/16 


4,350 


1/4 


4,600 


3/16 


9/16 


5,800 


7/16 


3/32 


5,630 


5/16 


6,100 


3/16 


23/32 


7,400 


9/16 


7/64 


7,560 


3/8 


8,500 


1/4 


3/4 


10,310 


9/16 


7/64 


10,100 


3/8 


12,500 


5/16 


7/8 


15,040 


21/32 


9/64 


14,750 


7/16 


17,500 


3/8 


1-1/32 


21,800 


3/4 


5/32 


20,300 


1/2 



*Based on quarter hard carbon steel of an ultimate tensile strength 
of 55,000 lbs. per sq. in. 
**Diameter of hole in eye is pin diameter + 1/64 in. 






e£ 






Wwmm — 



Swaged Tie Rods 



292 



Art. 176 



Miscellaneous Tables and Charts 



TABLE XX XIV 
PROPERTIES OF SWAGED TIE RODS 



D 






Size and 




Length of 


Section 


(minimum) 


Threads* 
per 1 in. 




Usable 
Thread 










(Swaged) 


Threaded 


Swaged 




.101 




Portion) 

.0080 


Portion 


Portion 


6—40 


.715 


1,150 


1,000 


10—32 


.134 


1.000 


.0141 


2,200 


1,900 


12—28 


.155 


1.063 


.0189 


3,000 


2,600 


1/4 —28 


.180 


1.219 


.0255 


4,000 


3,400 


5/16—24 


.223 


1.250 


.0391 


6,550 


5,700 


3/8 —24 


.274 


1.375 


.0590 


9,800 


8,500 


7/16—20 


.326 


1.500 


.0835 


13,200 


11,500 


1/2 —20 


.377 


1.625 


.1117 


17,800 


15,500 


*Star 


dard U. 


S. Threads. 













h L+M- 



Streamline Wires 



293 



Appendix 



Art. 176 



TABLE XXXV 
PROPERTIES OF STANDARD STREAMLINE WIRES 



B 



D 



Area of Cross Ultimate Strength in lbs. 



Size and 






Length of 


Section 


(minimum*) 


Threads* 
per 1 in. 






Usable 
Thread 












(Swaged 


Threaded 


Streamline 




.192 


.048 




Portion) 

.0071 


Portion 


Portion 


6—40 


1.0 


1,350 


1,000 


10—32 


.256 


.064 


1.1 


.0125 


2,700 


1,900 


1/4 —28 


.348 


.087 


1.3 


.0234 


5,000 


3,400 


5/16—24 


.440 


.110 


1.5 


.0376 


8,050 


5,700 


3/8 —24 


.540 


.135 


1.7 


.0563 


12,400 


8,500 


7/16—20 


.636 


.159 


1.9 


.0781 


16,700 


11,500 


1/2 —20 


.732 


.183 


2.0 


.1026 


22,800 


15,500 



*Standard U. S. Threads. 
M equals 1/2 in. for lengths less than 8 ft. For greater lengths, 
M must be increased 1/4 in. for each 1 ft. increase in length. 



TABLE XXXVI 
STRENGTH AND WEIGHT OF STEEL WIRE CABLE 





19 Strand; 


> Non-Flexible 


7x19 Extra-Flexible 






Cable 


Cable 


Diameter 
in 










Approx. 


Breaking 


Approx. 


Breaking 


inches 


Weight in 


Strength 


Weight in 


Strength 




lbs. per 


in lbs. 


lbs. per 


in lbs. 




100 ft. 


(minimum) 


100 ft. 


(minimum) 


375 3/8 




26.45 


14,400 


344 11/32 






22.53 


12,500 


312 5/16 


20.65 


12,500 


17.71 


9,800 


,281 9/32 






14.56 


8,000 


250 1/4 


13.50 


8,000 


12.00 


7,000 


,218 7/32 


10.00 


6,100 


9.50 


5,600 


,187 3/16 


7.70 


4,600 


6.47 


4,200 


,156 5/32 


5.50 


3,200 


4.44 


2,800 


,125 1/8 


3.50 


2,100 


2.88 


2,000 


.109 7/64 


2.60 


1,600 






.094 3/32 


1.75 


1,100 







294 



Art. 176 



Miscellaneous Tables and Charts 



o ♦> o a<. 

p. a * • 
* W f 

sets, 

Jsj. 

H g.R > • 

H«23 

list 






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tO •* •* 



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cm cm 

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295 



Appendix 



Art. 176 



TABLE XXXV III 
MISCELLANEOUS POWER PLANT WEIGHTS 



Curtiss 0X5 
LeRhone"110' 

Propeller -(-light wood) 20 

Propeller (heavy wood) .... 
Radiator (nose type) empty. 

Radiator water 

Water in piping 5 

Radiator Connections and 

Shutter 15 

2 Short Manifolds 

2 Long Manifolds 

I Set of Stacks 

Specific gravity of best test gasoline . . . 
Specific gravity of good test gasoline. . 

Weight of good test gasoline 



Hispano- 
Suiza E 

30 



55 
35 

5 

15 

23 



Hispano- 
Suiza H 

35 
55 
75 
45 
5-10 

15 
15 



Liberty 
12 

45 

60 

115 

65 
5-10 

25 

22 

15 



71 

72 

S 6.0 lbs./U. S. gal. 
' ' I 7.2 lbs./Imperial gal. 

Weight of engine oil " H Jbs./U. S. gal 

& s (9.0 lbs./Imperial gal. 

Weight of water 8.33 lbs./U. S. gal. 

Weight of plain sheet steel gasoline tanks, .60 to .80 lbs./U. S. gal. 

W T eight of leakproof gasoline tanks, 2.0 to 2.5 lbs./gal. for small gravity 

tanks, 1.5 to 1.8 lbs./gal. for 80 to 30 gal. tanks, and .8 to 1.2 lbs./gal. 

for 500 to 100 gal. tanks. 

Weight of standard storage battery for ignition system, 10.5 lbs. 

Weight of standard switch and voltage regulator for ignition svstem, 
2.4 lbs. 

Propeller, radiator, and radiator water weights do not vary ap- 
preciably with different types of engines, but with their horsepower. 

Contrary to last foot note to Table XXXVII the weights of standard 
storage battery and switch and voltage regulator are included in the dry 
weights for the three Packard engines, for the Hall-Scott L6-A, and 
for the Lawrance radial. 



296 



Art. 176 



Miscellaneous Tables and Charts 



TABLE XXXIX 

PROPERTIES OF STEELS 

U. S. Army Specifications 



- 










































ft 


SS2i 


fjSS fe§3SJ?S 




: : 252 


S3 : 




£■ 


ffOJITIO^ i. 








r J.r.lc. 


a c lM»rl..« - 


l»l»lo . 


S3 












• ■ 














































or -ola-rolllnK or ol 


-dialing. 






fc- : 

gS2 


sss; 


:g ;S gssaa- 


3« ssi 




« : S«. 


11: . 


fsrarir-i-rs; 










#§§§. 


! 1I1 iliili 


§38 -5gil 


:§§ IS?:= 


11 111 S 


" 


ID 3I.SOlflCatl3.16 10.2 


»-3 and 10, 


C«-A J-.6 2t.6..^h r. 

Ilp3. 




zt 


Ziii> 


,' 3 ° ssss^s 


gss ^SSS 


;J9 g? c g 

3 ' 


a? sss s 






„,,, t 








-snr»a 




























151 

sis 


liiii 


III Hilgs 


188 j||| 
sfsg gSSS 


III fill 

is; »sVg 


il in s 


§ 


10 2C0-3 42.000 
10.2CJ-A 66.000 


60]oOO 36|000 5o|oOO 






■s § 


> § 




s a 


s „ - * 




6s used d«p«Dd3 oo the 


jmtici -^ 


Jhloh ih. 5.rl is te 


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id 3 


























































I 


fil-i ! 


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a. 


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Jill- ri 


? id a.d .c- 

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z 




i S 


§ 


& 




Jura .J»s"to*.75 In. 


69.000 5C 


.000 13 39 






f,?,?,h. ■ 

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8§8 IHIsS 


^isl ^i§il 


h gigs 

JSS S2S2 


i« ,;L 


j 


3^1. )'.T 1.60 lr.. 


66,000 5C 


.000 15 36 






























T 






.'i".r 


Cr66D 


»«»» 


?t6ap!-.on>» 


Sulphur 


3.C4.1 


„,„,U. 


« 


. 


i 






1010 


. 16-0. 26 

.16-0.35 
.20-0.30 


.60-0. «) 


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.045 

.045 


0.050 

.060 
.06-. 05 






. T 


.._ _J___L 




« 


!« 


1016 


« 




. 


. 




1020 












•1020 






• 






1026 








. 






1090 


.30-f'.40 


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!c45 


.060 








• 


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• 


1035 






• "1 






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1046 


■;;;;; r 


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.46-0.55 
.50-0.60 

.O5-0.75 


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.-■6-0.50 
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.0~0 


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4 4-1 




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1066 










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! 1080 






■ v 






1 1096 






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3.26-3. 






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297 



Appendix 



Art. 176 



CO 



co 
O 

o 

CO 

S3 

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2 



SOLPOSI'i'lUl. 4 PaOKS!rii.3 0* - 

Nu.-Kl.uiUUJ ;JJ.'aL3 

air sraviOE, Ei;si;i'EHi::ci bivision. 






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298 



Art. 176 



Miscellaneous Tables and Charts 





ChAqt fog 


Lomo 5pquce Srpurs 


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I C = / for pin ends 






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Fig. 182. Chart for Long Spruce Struts (Euler's Formula) 



299 



Appendix 



Art. 176 



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t/Y (-*)?j^2*-j\ *c/ o/nujjoj 0//oqojod .?<// joj sjjo^/o s>t/j *& 
C9COJ uoncu/ ca/qoj. 21/4 0/ u9/i/6 sn/0/1 ?<// uot/j CCJ7 c/ 7 usum ja^Nj r/*j_ jcyy io/v o(J 
O Aq jjoc/o jc// uo u3ai6 d jo sn/OA pq/ A/cf/^/nuj /<j usq/y /=0 joj ps^o/d e/ /joqo r/qj 



oc/ou/ 

Mill 



V6u97 



U 






8 J? ^ 5^j§ 



<?> *» 









2 «! 









"* ;«. -s» -*> 



*F 



I ■ Im . l . 



4 



5 1 









Q_ <-> ° 



nrn 



r^ 



s-punO(j w poo 7 



|!|lii|llll|l!ll|IHI|llll|llll|llll|l!ll|ll 



spunoc/ jo zpuomouj u/ PO07 



300 



Art. 176 



Miscellaneous Tables and Charts 



Cn^er roe CaeBon Jteel, Spec/e/c#t/o/v 70,22-5 
p - \d 6,000 - ! Hf^ 6^] ^ FoePwfNDJ c-/ 



36, 
4c 
6 * £6,000,000% 



Fop F/xed Ea/ds c~4 

po/a/ts /?££ s ho whom th/s ch/9pt 
fop c=l c=2 amd c-J. The thpee po/nts fog amy omf 

TUBE APE CONNECTED BY A JOUD L/ME. Po/M7S FOG THE SAME 
VALUE OF C AeE COMMECTED //V OPDEP OF SECT/OMAL 
HP FA, 8Y DOTTED L/A/ES. 

THE TABLE SHOWS THE MAX /MUM LEM6TH FOG EACH 
TUBE, FOP IMH/CH THE CHAPT /S COPPECT. FOP 
OPEATEP LEA/OTHS THE CHAPT &/VES VALUES 
BELOW THE ALLOWABLE AMD EuLEP'S FOPMHLA 
SHOULD BE USED 

VALUES OF C BE/MO USED. 




P* CTr J £J THE SAME 



O.D. 


&Aoe 


May. L£MGTHS 




c~? 


c-s 


i " 


22 


mo 


294 


360' 




20 


EO.4 


26.9 


35.4' 


r 


22 


262 


37.0 


455' 




20 


260 


36.7 


45.0' 


3 ' 
4 


22 


J 1.7 


446 


55.0 




20 


3 IF? 


44.4 


544 




/6 


JO. 6 


43.6 


S3.S 


i' 


20 


36.6 


52/ 


64.0" 




fd 


36.3 


S/.4 


63.0 


/ ' 


20 


42.3 


60.0 


73.5' 




/6 


4/6 


59.0 


72.5' 




/6 


412 


594 


7/.S' 


'$' 


20 


476 


675 


630' 




16 


47.2 


66.9 


62.0' 




AS 


46.7 


66.0 


61.0' 


4 


te 


53.7 


74.5 


9/4' 




75 


622 


740 


90S 


i? 


Id 


63.6 


90.0 






i 


63.0 


691 




/f 


k 


74.0 








1 
3$ 


72.7 






i' 


16 


64.9 








* 


63.5 







-60 









— k "** 



k <* 

k io 
^ ki 



0s 



13 



*i 



Fig. 184. Chart for Carbon Steel, Specification 10,225 



301 



Appendix 



Art. 176 



'• \90, 



000 



m 



§£%?§ (p J# For Pin Ends C-l 

£-30,000000 /bs.per sg.i/2 For Fixed Ends c=4 

Points ore shon/n on this chorf for C = /,C= 2 , and ' C=3. The -three 
points for any one tube ore connected bg a so/id line. Points -for the same 
value of c ore connected, in order of sect/on a f orca, bg dotted tines . 
The table shows the maximum length For each tube, For which 
the chart is correct. For greater lengths the chart gives vo/ues 
below the o//owob/e and Eu/ers Forrnu/o should be used. P^ w 
the some vo/ues of c being used. 



~Z 




0.2). 


Gage 


Mox. Lengins 


C*l 


C-2 


c=3 


r 


22 


27.9 


394 


46.3 




20 


27.739-2 


46.0 




Id 


27.3 


36.6 


47.3 




i 
16 


27.9 


36.1 


46.7 


^ 


20 


31.3 


442 


54.1 




IS 


30.9 


43.7 


535 




i 

16 


30.5 


43.2 


530 


'* 


20 


34.6 


49.2 


603 




/6 


34.5 


46.7 


59.6 




/ 

10 


34./ 


46.2 


590 


ii 


20 


42.0 


59.4 


726 




i 
16 


41.3 


56.3 


7/5 


/# 


20 


49.2 


69.5 


63.3 




/ 
16 


46.4 


66.5 


639 




-3 

32 


476 


673 


32.5 


2 


20 


36.4 


7.9.6 


97.6 




) 
16 


36.6 


76.6 


96.4 




32 


S4.7 


77.4 


94.6 



Chorf for f9//og Steel Tubes, Specif /cation /0,2$7 



Fig. 185. Chart for Alloy Steel Tubes, Specification 10,227 

302 



Art. 176 



Miscellaneous Tables and Charts 



Chart Foe Dupalc/m/h Tubes 

P • U7000- -£Z^?7-)'V For p,n ends C-/. 

5-10, 000 t 000 tbs. per ja. in For "fixed 'ends C~d. 
Points ore shown /n th/s chart for C-t, C-Z, and C=3. The 
three points -for ony one tube ore connected by o so//d t/ne 
fb/nts for the some ya/ue of C ore connected /n order of 
oreo, by dotted t/nes. 
The tab/e shows the rnox/'mum /enyth for eoch it/he for 
**vh/ch the chort as COr reed For yr eater tenytbs the chart 
g/ves yo/ues be tow the at/owobie ond fetters Forma/a 
shou/d be used. P - Q2Z 7e~~ where c is the s-orne as- 

obo^e For the fu/er ronye the chort For s/eei tubes' 
> \used os tony CO turn ns moy be used, d/v/d/'ny 
the vo/ue of P obto/ned by 3 




oosem 



0042290 



0053 285 



P07CS5.7 SO J 



ooirsae 



o./zs 






433 

6 



SOZ 



007ZS34 8Z.6 



0093 S7 8 81 7 tOQO 



007Z 



Lenyth 



C-Z C=3 



30<t\37.? 



404 49.5 



S/O 6Z6 



6/8 



6Z0 



6/Z 750 



7ZS869 



7/. 6 



880 



7, 



735 '040 >Z73 



7Z0 'O/.7/ZS.0 



87/ 

/0/Z 



Fig. 186. Chart for Duralumin Tubes 



303 



INDEX 



Page 

Aerofoils, Characteristics of 34 

Selection for Best Cruising Radius 31, 32 

Ailerons, Calculation of Moments 177 

Aileron Lever, Investigation of 184, 185 

Aileron Torque Tube, End Thurst Bearing for 193 

Investigation of Teeth in Collar on . 187 

Investigation of Attachment of Masts to Tubes 190 

Air Controls, Relation of, to Pilot 197 

Airplane, Characteristics of Military Airplanes 41, 42 

Analytical Determination of Deflection of Beams 277, 278 

Analytical Solution for Truss Stresses 4 

Ash, Commercial black, Properties of 288 

Commercial white, Properties of 288 

Aspect Ratio, Consideration of visibility, structural weight, aero- 
dynamic efficiency and stability 44 

Effect of varying 37 

Axle, Design of 169 

Factor of safety for 169 

Loading conditions 169 

Specifications for axle steel 169 



Balance Drawing 231 

Balance Table 230 

Basswood, Properties of 288 

Beams, Built-up 134 

Calculation of horizontal shear stresses 17 

Calculation of moments and shears for simple beams 12 

Beam Theory, Assumptions and limitations of . 11 

Development of 9 

Bell Crank Lever, Investigation of 185 

Stress on weld between lever and torque tube 186 

Bending, Combined bending and compression 22 

Formula for wood and steel in combination 24 

Formula for combined torsion and bending 26 

Birch, Properties of 288 

Black Cherry, Properties of 288 

Blind Area, Methods of eliminating 29 

Bolts, Areas of 138 

Design of 138, 139 

Horizontal vs. vertical bolts 140, 141 

Brash, Definition of 284 

304 



Index 



Page 

Cabane Struts and Wires 129 

Cable, Strength and weight of steel wire cable 294 

Capstrips, Split 112 

Cedar, Port Orford, Properties of 287 

Ceiling, Density 31 

Center of Pressure Movement 61 

Center Section Cross Wires and External Drag Wires, Capacities of 132 

Center Section Struts, Analysis of 128 

Loading Condition 128 

Loading producing limiting stresses 131 

N Tvpe 129 

Tripod Type 129, 130 

Unstability of 131 

Chassis, Bracing wires 145 

Characteristics of chassis structure 144, 145 

Conditions of loading 143, 144 

Connection of chassis strut to longeron 144 

Connection of spreader tubes to guide plate 144 

Design, Example of reduction of eccentricities 168 

Factors of safety 168 

Relative strength of members 168 

Requirements of 167 

Streamlining 168 

Dynamic factor for 144 

Eccentric moments 145 

Example, Actual stresses and moments in members 159 

Calculation of moment due to rise of axle 163 

Calculation of moments perpendicular to plane of struts caused 

by rise of axle 162, 163 

Computation of projection of members 147 

Data used in 145, 147 

Distribution of moments in a vertical plane perpendicular to 

plane of struts 153, 154 

Distribution of moments between members of chassis, 153, 154, 155 

Distribution of moments in the plane of the struts 158 

Introduction of method oi index loads 147, 148 

Introduction of principle of least work 152 

Moments in struts, Application of principle of least 

work 153, 154, 155, 156, 157 

Statically determinate condition, side thrust of unity perpen- 
dicular to axis 150 

Thrust of unity parallel to axis 150 

Vertical load of unity, one-wheel landing 149 

Vertical load of unity, level landing 148, 149 

Staticallv indeterminate condition, vertical load, one-wheel 

Ending 150, 151, 152 

305 



Index 



Chassis, Example — Continued Page 

Statical solution 148 

Stresses considered 147 

Use of determinates 149, 150 

Factors of safety for . 175 

General procedure in selecting loading conditions for investi- 
gation 144 

Hinged axle 165 

Loading, Limiting case for diagonal wires 144 

Limiting case for front strut 144 

Limiting case for rear strut and diagonal wires 144 

Material used in 145 

Members, Eccentricities of 145 

Relation to fuselage 176 

Spreader Tube, Attachment of, to struts or guide plate 144 

Calculation of 163 

Loss in strength due to brazing 167 

Cherry, Black, Properties of 288 

Clark Truss 98 

Climb, Rate of 31 

Cockpit, Comparative dimensions of 197, 200, 201 

Column curve for spruce struts , 282 

Column formulas 23 

Column formulas for struts 122 

Columns, Euler's formula 23 

Formulas for wood and steel in combination 24 

Johnson's parabolic formula 24 

Combined bending and compression 280 

Combined spiral and diagonal grain, Effect of 283 

Compression, Combined bending and compression 22 

Continuous beams, Equations for 259 

Controls, Curved rudder bar 177 

Control Design, Calculation of moments 177 

Data for 177 

Dual stick 193 

Loading 177 

Control Mast, Attachment to surface 204, 205 

Connection of mast to tube 204, 205 

Hollow 205 

Plywood mast, with steel core 205 

Relation to control cable 205 

Steel core with wood fairing 205 

Controls for large airplanes, Relative advantages of various types. . 197 

Controls, Relation of air controls to pilot 197 

Control stick, Column action in 183 

Investigation of, for elevator loads 180 

Investigation of, for aileron loads 183 

Controls, Stress analysis for 177 

306 



Index 



Page 

Control Surfaces, Analysis of braced 205, 226 

Analysis of unbraced 223 

Auxiliary spar 210, 211 

Bracing 205 

Double cambered 209, 221 

Elevator hinge 217 

General considerations 204 

Government specification loading for \ . 223 

Joint between torque tube and ribs in tubular construction. .218, 220 

Stabilizer box rib 211, 212 

Strength and weight of 206 

Stresses in braced rudder post 227 

Tabulation of data for 206 

Tubular construction of 218 

Control System, Allowable unit stresses 202 

Basis of design 177 

Failure of . . 177 

For pursuit airplane 193 

Minimum strength of 177 

Relative strength of component parts 177 

Controls, Torque system of control 177 

Control, Torque tube connecting stick in dual control system 193 

Deflections in continuous spans 259 

Deflection moments in spars 64 

Deflection of beams, Methods of determining deflections of beams 

with varying load and section 270 

Deflections, Primary and secondary deflections in spars 67, 68 

Determinants, Application of 149, 150, 151 

Diagonal grain, definition of 283 

Effect on strength properties of wood of 283 

Diving, Stresses produced in 57, 58 

Douglas Fir, Effect on strength of spiral and diagonal grain.... 286 

Properties of 285, 287 

Drag, Resolution of lift and drag forces 63 

Drag truss, Solution of stresses in 107 

Drag trusses, Stresses in 63 

Dual stick control 193 

Dual stick control, Torque tube between sticks 193 

Duraluminum, Properties of 289 

Dynamic factor 230 

Eccentricities of fittings 53 

Eccentric moments and shears, Division among spars of 106 

Eccentric strut loads and wire pulls 64 

Elastic curve of beam 270, 271 

307 



Index 



Page 

Elevator cables, Stresses in 196 

Elevators, Calculation of moments 177 

Elevator tubes, Investigation of mast on 189 

Elevator load , Calculation of center of gravity 177 

Elevator, One mast 177 

Elevator hinge 217 

Elm wood, Properties of 288 

Empennage, Air loads 234 

Example of internally braced construction 205 

Strength factors for 256 

Engine weights and horsepower, Data on 295 

Engine, Torque of 51 

Equations for continuous beams 259 

Equilibrium, Equations of 1 

Fineness Factor, Definition of 256 

Elements determining value of 256 

Ranges of 256 

Fineness Ratio 128 

Fir, Douglas, Properties of 287 

Fittings, Eccentricities of 53 

Design of 142 

Wing spar 83 

Footpad, Adjustment of 197 

Forces, Characteristics of 1 

Force polygon, Construction of ." . 8 

Free body method 7 

Funicular polygon 271, 272, 274 

Fuselage, Air loads on horizontal tail surfaces 234 

Balance drawing 231 

Balance table 230 

Computation of factors of safety 250 

Conditions of loading assumed for stress analysis . 229 

Degree of fixity of longerons 253 

Discussion of factors of safety 252 

Distribution of torque load between reactions 235, 236 

Distribution of rudder air load between the upper and lower 

truss 237, 239 

Division of weight of fuselage and equipment between panel 

points 232, 233 

Dynamic factor 230 

Inertia load 243 

Load Factor 230 

Method of compensating for the displacement of the center of 

gravity 236, 237 

Monocoque type 228 

Advantages of 229 

308 



Index 



Fuselages, Monocoque Type — Continued Page 

Allowable unit stress in compression for plywood 255 

Analysis of stresses 255 

Construction of 228, 229 

Disadvantages of 229 

Nature of failure 255 

Secondary compressive stresses 255 

Properties of sections 248, 249 

Plywood covered type 228 

Relative advantages of various types of 228 

Semi-monocoque type 228 

Action of bulkheads 254 

Analysis of stresses 254 

Application of loads 254 

Construction of 229 

Material used in 229 

Stick and wire type 228 

Strength factor for 256 

Stress analysis of 229, 230 

Stresses for flying condition 237 

Stress analysis for flying condition with rudder turned 237 

Stress analysis for condition of landing with tail up 240 

Stress analysis for flying condition with air loads on horizontal 

tail surfaces 234, 235 

Stress analysis for three point landing condition 243 

Summary of stresses in fuselage 245 

Three point landing condition, Distribution of load between the 

chassis and tail skid 243 

Torque, Calculation of 235 

Traction, Calculation of 235 

Truss type 228 

Veneer covered truss type 253 

Welded metal fuselage 228 



Gap-chord Ratio, Effect of varying 37 

Graphical solution of deflection in beams 270 

Graphical solution of truss stresses 7 

Graphical solution of truss stresses, sign of stresses 9 



High incidence condition 55 

High incidence, Load factor for 258 

Hooke's Law 1 1 

Horizontal shear, Strength value of wood in 281 

Horsepower of engines 295 

309 



Index 



Page 

Incidence, High and low incidence condition 55 

Inertia load 243 

Interplane struts, Location of 47 

Joints, Analytical method of 4 

Laminated struts 124 

Landing condition, Analysis of stress, Tail up 240 

Landing speed, Expression for determination of 30 

Least work, Application of method of 152, 154, 155 

Method of 52, 152 

Lettering of truss 7 

Lift, Resolution of lift and drag forces 63 

Lift truss, Analysis of stresses in 81 

Clark type 98 

Division of loads from lift trusses 106 

S.P.A.D. type 100 

SA'.A. type 100 

Stresses in 63 

Thomas-Morse type 100 

Lift wires, Location of, in inner bay of airplane 45 

Load, Distribution of load between spars 60 

Load factor for fuselage 230, 256 

For tail surfaces 256 

For wings 256 

Load on wings, Distribution of load along span and chord 60 

Maximum values of 54 

Loads from the lift trusses, Division of 106 

Loading, Rib 117 

Longerons, Splices in spars and longerons 137 

Low incidence, Load factor for 258 

Lugs, Properties of standard lugs 292 

Lugs and eye-bolts, Design of 139 

Maple, Properties of 288 

Mast, Investigation of mast on elevator tubes 189 

Metal, Properties of non-ferrous metals 298 

Member, Two force 2 

Military Airplane, Characteristics of 41 

Modulus of Elasticity of Wood 280 

Modulus of Rupture of Routed Sections 279 

Modulus of Rupture of Spruce 280 

A4oments, Calculation of, For Aileron 180 

For Aileron, Dual Stick Control 193 

For Aileron Lever 1 84 

For Continuous Beams 19 

310 



Index 



Moments — Continued Page 

For Elevator 1 80 

For Elevator, Dual Stick Control 196 

For Rudder 180 

For Simple Beams : . 12, 13, 14 

Deflection 64 

Diving 57, 58 

Plotting of moment and shear curves 15 

Relation between moments and shears 15 

Theorem and equation of three moments 18 

Monocoque Type of Fuselage (See fuselage) 228, 255 

Monoplane Wings 100, 103 

Motor, Power of 30 

Multi-spar Construction, Distribution of load between spars 103 



Newton's First Law of Motion 1 

Nomenclature for equations of continuous beams 259 

Non-Ferrous Metals, Properties of 298 



Oak, Properties of 288 

Outboard Strut, Use of 276 



Parachute Pack, Standard dimensions of 197 

Pine, Eastern white, Properties of 287 

, Western white, Properties of 287 

Pins, Areas of 138 

Design of 138, 139 

Pin Plate, Design of 141 

Pin jointed truss 2 

Plotting of moment and shear curves for simple beams 15 

Plotting, Sign of shear for continuous beams in plotting 18, 19 

Plywood covered fuselage 228 

Plywood truss ribs, Reinforced 115 

Plywood web ribs 109 

Poplar, Properties of 288 

Power of Motor, Expression for determination of 30 

Power Plant Weights, Miscellaneous 296 

Prediction of High Speed at Ground 256 

Propeller Traction 235 

Propeller Torque 235 



Rate of climb 31 

Reactions, Calculation of 3 

311 



Index 



Reactions — Continued Pa ^ e 

Characteristics of 2 

Determinate and indeterminate 2 

Reinforced plywood truss rib 115, 116 

Resolution of lift and drag forces 62, 63 

Rib, Attachment of rib to spar 110 

Rib loading for testing 117 

Rib spacing 118 

Ribs, Assembly of Ill 

Compression 112 

Plywood web type < 109 

Plywood web type, Direction of grain 109 

Economical chord length of 109 

Material used in 110 

Recommended section 109 

Strength of 110 

Reinforced plywood web 115 

Split capstrip for . 112 

Stress analysis of 224 

Truss 113 

Tenoned web 110 

Thick wing section Ill 

Truss type, Section of 114, 116 

Strength of 113, 115 

Weight of 113, 115 

Veneering covering of 118 

Routing, Reduction in modulus of rupture of wood due to 279 

Rudder bar, Duraluminum base for 192, 193 

Investigation of 188 

Rudder, Calculation of moments 179 

Internally braced construction 214, 215 

Rudder post, Stresses in braced rudder, post 227 



Sections, Method of, For solution of truss stresses 7 

Semi-monocoque type of fuselage (See fuselage) 229, 254 

Shear, Method of 5 

Plotting of shear diagram 13, 15 

Stresses due to vertical shear 18 

Shears, Calculation of -. 14, 15, 19 

Relation between moments and shears 15 

Shock Absorber, Comparison of typical load elongation curves . . 171, 172 

Data necessary for computations 172 

Determination of number of strands of cord 173 

Factors of safety for 175 

Function of 170 

Government specification for 174 

312 



Index 



Shock Absorber — Continued Page 

Initial tension in cord 173 

Ranges of maximum elongation . . 173 

Types of 169 

Sign of Stresses in graphical solution 9 

Signs, Determination of signs of shear in three moment equation. . 18 

Law of 1 

Spar Sections, Properties of 52 

Spars, Deflection in 64 

Determination of moments, shears and reactions 63, 78 

Duraluminum 135, 136 

Location of 50 

Plywood web 135, 136 

Splices in spars 137 

Speed, Landing 30 

Prediction of high speed at ground 256 

Speed Range 30 

Spiral Grain, Definition of 283 

Effect on strength properties of wood of 283, 286 

Splices in spars and longerons 137 

Spruce, Strength values for 280 

Stabilizer, Internally braced construction 205, 206, 212 

Stagger, Effect of varying 37 

Static Test conditions for the wing cellule 56 

Statical moment, Calculation of 17 

Statics, Principles of 1 

Steel, Chart for alloy steel tubes 302 

Chart for carbon steel 301 

Chart of steel tubes 300 

Column curve for mild steel 25 

Properties of 297 

Wood and steel in combination 24 

Stick and wire type of fuselage (See fuselage) 228 

Streamline strut section 120 

Streamline wire, Properties of standard streamline wires 294 

Strength factors for fuselage 258 

For tail surfaces 258 

For wings 258 

Strength of Materials 9 

Strength properties of wood 281 

Stresses, Allowable unit stresses for members of control system. . . . 202 

Stress analysis for controls 177 

Stress analysis of braced control surfaces 226 

Of fuselages, monocoque type 255 

Of fuselages, semi-monocoque type 229, 254 

Of fuselages, truss type 229 

Of lift truss : 81 

Of rib 224 



313 



Index 



Page 

Of unbraced control surfaces 223 

Stress table for spars 65 

Stresses, Character of 5 

Determination of sign in graphical solution 9 

Diving 57, 58 

Horizontal shear stresses in beams 17 

In lift and drag trusses 63 

Unit fiber stresses 56 

Struts 119 

Cabane struts and wires 129 

Center section (See Center Section Struts) 128 

Chart for long spruce struts 299 

Column formulas for long struts 122 

Column curve for mild steel , 25 

Composite strut sections of steel and wood 25, 125 

Curves for compressive strength of spruce struts 282 

Design of 122 

Design of center section struts 122, 125 

Equivalent weight of 128 

Equivalent weight of, a minimum 128 

Fineness ratio for 128 

Hollow section, Properties of 120, 122 

Limiting dimensions of 124 

Laminated 124 

Location of interplane struts 47 

Resistance of 127 

Solid section, Properties of 119 

Streamline strut sections 119 

Subjected to bending about both axes 125 

Taper of 125 

VD correction factor for 128 

Tail Surfaces, Calculation of unit loads 179 

Government specifications for 177 

Strength factors for 258 

Tenoned web ribs 110 

Three-moment equation 18 

Tie Rods, Properties of swaged tie rods 293 

Torque of engine 51 

Torque, Propeller 235 

Torque tube carrying elevator masts and rear stick 195 

Torque tube, to which control stick is secured, Investigation of 184 

Torsion, Formula for combined torsion and bending 26 

Formula for simple torsion 26 

Torsional stresses, Reduction of torsional unit stresses in thin walled 

tubes 202, 225 

Traction, Propeller 235 

314 



Index 



Page 

Truss, Analysis of stresses in lift trusses 81 

Clark lift truss 98 

Definition of 1 

Determinate, indeterminate and unstable 2, 3 

Lettering of 7 

S.P.A.D. lift truss 100 

S.V.A. lift truss 100 

Solution of stresses in drag trusses 107 

Stresses in drag trusses 63 

Stresses in lift trusses 63 

Thomas-Morse lift truss 100 

Truss Stresses, Calculation of, by method of joints 4 

Calculation of, by method of moments 5 

Calculation of, by method of shear 5 

Truss type of fuselage (See fuselage) 228, 253 

Tubing, Hollow circular tubing 120 

Hollow elliptical tubing 122 

Properties of circular tubing 290 

Veneer covering for fuselage . 253 

Veneer covering of ribs 118 

Visibility, Importance of good visibility 28 

Methods of securing good visibility 28 

Vought VE-7 two-seater training airplane, Stress analvsis of fuselage 

of 229 



Warren truss, Calculation of reactions 3 

Weight analysis of airplanes 41, 43 

Weight of airplanes per horsepower 41, 44 

Per sq. ft. of wing area 41, 44 

Walnut, Black, Properties of 281, 289 

Welded metal fuselages 228 

Wheel control 197, 202 

Wing cellule. Static test condition for 56 

Wing sections, Characteristics and ordinates of 38 

Choice of 29 

Wing selection, Classification of Airplanes for Wing spar fittings. . 36 

Wing Stress Analysis, Application of the principle of least work. .93, 94 

Calculation of air load drag 95 

Calculation of drag components of struts and wires 89 

Calculation of moments and shears for eccentric wire pulls. ... 85 

Calculation of moments and shear for offset strut loads 83 

Calculation of moments and shears in spars 78 

Calculation of stresses in front lift truss 64 

Data for calculation of drag 89 

Deflection of front upper spar 89 

315 



Index 



Wing Stress Analysis — Continued Page 

Dimensions of struts 95, 97 

Distribution of loads between spars 70 

Distribution of loads between wings 70 

Distribution of pressure on wing tips 75 

Effect of routing on strength of spar sections 98 

Load on wings 75 

Load per inch of run 75 

Loads on drag trusses 95 

Location of center of pressure 70 

Properties of spar sections 87, 88 

Properties of strut sections 97 

Table of moments 81 

Table of moments, shears, and reactions for front upper spar. ... 86 

Wings, Distribution of load between wings 60 

Maximum values of load on 54 

Monoplane 100 

Strength factors for 258 

Wing tip, Distribution of pressure along wing tip 61 

Wing tips, Efficiency of various forms of 44, 46 

Wires, Capacities of center section cross wires and external drift 

wires 132 

Capacities of incidence wires 133 

Location of lift wires in inner bay 45 

Properties of standard streamline wires 294 

Wood, Brash . 284 

Effect on strength properties of spiral and diagonal grain 283 

Properties of, at 10 per cent moisture 278, 281 

Properties of wood in terms of properties of spruce 285 

Relative merits of 284 

Wood and Steel in combination 27 

Worm and gear control 197, 202 



316 



CO 



# 



^0 ^ 



